Chapter 6 Bodies in Equilibrium Fundamentals of Physics: A Senior CourseSHORT ANSWER 1. Several children are playing on the smooth, frictionless, level floor of a nursery. They hold strings tied to a small toy truck, which they pull horizontally. Barbara (B) pulls with a force of 16 N [east], Scott (S) pulls with a force of 19 N [west]. With what force must Kevin (K) pull to keep the truck in equilibrium, horizontally? ANS: REF: K/U MSC: SP OBJ: 2.2 LOC: FMV.02 KEY: FOP 6.1, p.205 2. A guy wire attached to the top of a gate post is anchored to the bottom of the next post, as illustrated. Assuming that the gate post does not move, what are the horizontal and vertical components of the force exerted on the top of the post by the guy wire, if the tension in the wire is 500 N? ANS: REF: K/U MSC: P OBJ: 2.2 LOC: FMV.02 KEY: FOP 6.1, p.208 3. A cyclist applies a downward force of 65 N on the pedal of her bicycle, as illustrated. Find the torque for each position. (a ) (b) (c ) (d) ANS: (a) (b) (c) (d) REF: K/U KEY: FOP 6.3, p.215 MSC: SP 4. Calculate the torque of each force about the corresponding point P. (a ) (b) (c ) (d) ANS: (a) (b) (c) (d) REF: K/U KEY: FOP 6.3, p.219 MSC: P 5. A uniform, 12.0 m long girder has a mass of 500 kg. It rests unattached, on a concrete slab with one end overhanging the edge by 5.5 m. How far can an 80 kg man walk out on the girder before it tips? Determine the tension in the guy wire. inclined 60° to the horizontal. and exerts a force of 600 N on the top of the mast (see diagram). p. The mast is pivoted on a hinge pin at its base.219 MSC: P 6. ANS: Taking torques about P: .ANS: Taking torques about P: REF: K/U KEY: FOP 6.0 m below the top of the mast. The mast is supported by a guy wire running to the ground from a point 6.3. A short-wave aerial is attached to the top of a mast 20 m high. REF: K/U. . The centre of gravity of a person may be located by the arrangement shown in the diagram. located from the left end? ANS: Taking moments about B: REF: K/U KEY: FOP 6. where is the centre of gravity of the system.219 MSC: P 7. MC KEY: FOP 6.4. locate the centre of gravity of the person. p. as illustrated. Assuming the scales to be adjusted to zero with the plank alone. p. If each block has a mass of 10 kg and a width of 24 cm.3. A 12 kg plank has four concrete blocks placed on it.223 MSC: P 8. 4. p. ANS: . REF: K/U KEY: FOP 6. given the locations of the centres of gravity of the upper and lower leg. Locate the centre of gravity of an extended leg relative to the sole of the foot. as shown in the diagram.ANS: Taking moments about A: Therefore the centre of gravity is located 1.2 m from his feet.223 MSC: P 9. p.0 cm by 20 cm.Taking moments about P: REF: K/U KEY: FOP 6.224 MSC: P 10.4. The centre of gravity for her upper body is located 70 cm from her hips.224 MSC: P 11. A 60 kg woman bends 90° forward from the waist. What is the tipping angle for a 2. how far forward from the line of her legs is her new centre of gravity? ANS: Taking torques about the centre of gravity: The centre of gravity is 47 cm from the line of her legs.0 cm by 12. REF: K/U KEY: FOP 6. keeping her legs vertical. p.0 kg brick.4. If two-thirds of her mass is in her upper body. 6. when it is placed on end? . Using the following diagram of a brick. (a) if it is standing on its side (b) if it is standing on its face ANS: (a) .227 MSC: SP 12. as illustrated. its centre of gravity will be located at its geometric centre. Also.5. it follows that Note that the mass of the object has no effect on the value of the tipping angle. find the critical tipping angle for rotation about an edge for the following. REF: K/U KEY: FOP 6. would be different if we tipped the brick on another edge. Of course. The only important criteria are the dimensions of the object and the location of the centre of gravity. p. the centre of gravity will be directly above the edge P.ANS: Since the brick is a uniform object. Given the dimensions in the diagram. at the critical tipping angle. 5. What is the critical tipping angle for the table illustrated? ANS: .(b) REF: K/U KEY: FOP 6. p.230 MSC: P 13. is pushed at its peak. what angle will the base of the triangle make with the horizontal when it just begins to tip over? ANS: Centre of gravity is located at or 2. p.REF: K/U KEY: FOP 6.5. .230 MSC: P 14. If the triangle is pushed parallel to its base. with equal sides of 10 cm. A triangular steel sheet.9 cm from the base. Note: Symmetry alone would tell us this.5 LOC: EM1. p.6. p. What is the mass of a fish that stretches the spring 7.08 KEY: FOP 6. A spring in a pogo stick is compressed 12 cm when a 40 kg boy stands on the stick.5 cm from its normal length? ANS: REF: K/U MSC: P OBJ: 4.5 LOC: EM1.6. p.5.6.232 16.232 .08 KEY: FOP 6. What force is necessary to stretch a spring whose force constant is 120 N/m by an amount of 30 cm? ANS: REF: K/U MSC: P OBJ: 4. p. REF: K/U KEY: FOP 6.5 LOC: EM1. A spring with a force constant of 600 N/m is used in a scale for weighing fish. What is the force constant for the pogo stick spring? ANS: REF: K/U MSC: P OBJ: 4.08 KEY: FOP 6.232 17.230 MSC: P 15. If the column supports a load of 5. how much is it compressed? ANS: REF: K/U KEY: FOP 6. a mass is attached to the foot. frictionless surface if pulled by a spring with a force constant of 40 N/m. as illustrated.7.4 kg.08 KEY: FOP 6. A rectangular steel column 2.5 × 103 kg.7 × 10–4 m2. During knee exercises.5 LOC: EM1. stretched by a constant amount of 8.238 MSC: P 20. . What is the acceleration of a 2. p. p. with a cross-sectional area of 2.232 19.0 cm? ANS: REF: K/U MSC: P OBJ: 4.0 kg cart on a flat. is used vertically to support a sagging floor. The force applied to a dynamics cart is measured with a stretched spring.5 m long. if the mass of the lower leg and foot is 5. Find the torque exerted on the knee joint by the leg and mass.18.6. ANS: REF: K/U KEY: FOP 6.9. The deltoid muscle raises the upper arm to a horizontal position.249 MSC: P 21. p. find the following. (a) the tension in the deltoid muscle (b) the components Rx and Ry of the force exerted by the shoulder joint (c) the magnitude of the resultant force on the shoulder joint ANS: .1 kg and its centre of gravity is considered to act 36 cm from the shoulder joint. as illustrated in the diagram. If the mass of the arm is 4. 249 MSC: P 22. p.0 m? ANS: .(a) Taking torques about P: (b) Taking horizontal components: Taking vertical components: (c) REF: K/U.80. MC KEY: FOP 6.9. If the coefficient of friction between your running shoes and the gym floor is 0. how fast could you run around a circle with a radius of 2. 02 KEY: FOP 6. MC MSC: P OBJ: 2. A guy wire makes an angle of 55° with the ground.02 KEY: FOP 6.256 24. Find the vertical and horizontal components of the force exerted on the pole by the wire.REF: K/U MSC: P OBJ: 2. .4 LOC: FMV.2 LOC: FMV. If the arrow makes an angle of 55° with the string above and below the arrow.256 25.11. and applies a force of 1600 N to the top of a vertical clothes-line pole. p. p. An arrow is drawn back against a bowstring with a force of 280 N. You are wearing shoes that have a coefficient of friction on rock of 0.70.02 KEY: FOP 6. On how steep a rock slope could you stand without slipping? ANS: REF: K/U.11.4 LOC: FMV. find the tension in the string. ANS: REF: K/U MSC: P OBJ: 2. p.11.256 23. A tennis net is supported at each end by a steel post.) ANS: Horizontal components: Vertical components: . as illustrated.11.01 KEY: FOP 6. calculate the tension in the guy wire and the compression in the steel post. (Assume a hinge pin at the base of each post. If the maximum tension in the net rope is 800 N and the angles are as shown. p.2 LOC: FMV.256 26.ANS: REF: K/U MSC: P OBJ: 2. p.256 27. the angles at the pulley on the foot can be changed.02 KEY: FOP 6. The tendons pull on the bones of the lower leg with a total force of 100 N. p. as illustrated. Would the angles be made larger or smaller. (c) If they were not equal.11.2 LOC: FMV. to decrease the force on the leg? Explain your reasoning. When the femur is healing. REF: K/U. The Russell traction apparatus is used for a fractured femur. (c) Why are the angles kept equal.02 KEY: FOP 6. the sole of the foot).e. The Russell apparatus provides a force parallel to the femur that prevents contraction of the femur. (a) Find the magnitude and direction of the horizontal force on the leg (i.257 28. (b) To adjust the size of this force. as shown? ANS: (a) (b) The larger the angle. which could result in a shortened leg after healing. MC MSC: P OBJ: 2. the lower the force.11. The force would not be along the length of the leg as required. there would be a vertical net force and the foot would be lifted..2 LOC: FMV. The sling supports the weight of the lower leg.REF: K/U MSC: P OBJ: 2. the muscles of the upper leg tend to pull the two sections of the femur together. What torque is exerted on the leg by the tendons alone? . 5 m from the pivot. . The other is seated 2. is seated 2.ANS: REF: K/U KEY: FOP 6. Two children balance on a seesaw.11. p.257 MSC: P 29.257 MSC: P 30. with a mass of 30 kg. Find the tension in the bicycle chain shown.0 m from the pivot. p. What is the mass of the second child? ANS: Taking moments about P: REF: K/U KEY: FOP 6. The bottom is slack.11. One. ) REF: K/U KEY: FOP 6. REF: K/U KEY: FOP 6.257 MSC: P 32.11. p. What force must the hand provide perpendicular to the pole? .ANS: Net torque about axle of crank is 0.11.257 MSC: P 31. What total vertical force is required to lift the handles of the loaded wheelbarrow illustrated in the diagram? ANS: Taking moments about P: (The force per handle would be 105 N. A marcher in a parade holds a flag at an angle of 60°. p. 258 MSC: P 33. Three men. p.mass of pole and flag = 3.8 kg ANS: Taking torques about P: REF: K/U KEY: FOP 6. Two are at one end of the pole and the third is at the other end. are carrying a heavy load on a pole.0 m apart as illustrated.11. 4. Where should the load be placed so that each man gives equal support? ANS: . ANS: (a) Taking moments about P: .44 kg.Taking vertical components: Taking moments about P: The load should be placed 1. REF: K/U KEY: FOP 6.0 – 1.11. The diagram shows a forearm supporting a shot put whose mass is 5. p.3 = 2. find the force exerted by the biceps and the force on the elbow joint. or 4. Using the dimensions on the diagram.7 m from the single man.3 m from the end closest to the two men.259 MSC: P 34. (b) Taking vertical components: REF: K/U KEY: FOP 6. To make larger pictures. an enlarger can be moved to the edge of a table so that the image is projected onto the floor. what will be the reaction force at B? mass of base = 2.259 MSC: P 35.0 kg ANS: (a) Taking torques about B: . (a) What is the minimum mass required on the base at point A to keep the enlarger from tipping off the table? (b) With this mass in place.11. as shown.0 kg mass of enlarger = 6.0 kg mass of support = 2. p. (b) Taking vertical components: REF: K/U KEY: FOP 6.11, p.260 MSC: P 36. Earth and the Moon are 3.8 × 108 m apart, and their masses are 6.0 × 1024 kg and 7.3 × 1022 kg, respectively. Where, approximately, is the centre of gravity (relative to Earth) of the Earth–Moon system? ANS: Imagine a force “supporting” Earth and the Moon in balance. To give some meaning to this supporting force, assume a gravitational field g due to the Sun, located perpendicular to the Earth–Moon axis. Taking torques about E: Note that the centre of gravity lies inside Earth! REF: K/U KEY: FOP 6.11, p.261 MSC: P 37. The four-legged table shown has a mass of 30 kg. Assuming that the legs are massless and the table does not slide, at what angle will it tip over? ANS: REF: K/U KEY: FOP 6.11, p.261 MSC: P 38. A tall water glass with a mass of 300 g is about to be pushed (horizontally at its rim) across a table. The glass has a base 1.8 cm diameter and is 6.0 cm tall. If the coefficient of friction with the table surface is 0.20, determine whether the glass will slide or tip. ANS: Taking torques about P: Since the glass will tip. This is the minimum force to tip the glass. But the minimum force to slide the glass on the horizontal table is the maximum frictional force calculated as follows: Hence the glass will tip and not slide when REF: K/U KEY: FOP 6.11, p.261 . MSC: P 39. A 50 kg high wire artist carries a weighted pole, as illustrated. Her centre of gravity is 80 cm above the high wire, and the two 23 kg masses are located 120 cm vertically below the wire. Find the vertical location of the centre of gravity of the walker and the weighted pole, relative to the high wire. Is this a case of stable or unstable equilibrium? ANS: REF: K/U KEY: FOP 6.261 MSC: P 40.11. when he is carrying a 14 kg pail of water. Find the minimum lateral shift in position of the centre of gravity of a standing man. p.Taking moments about T: She is stable since the centre of gravity lies on a line with her feet. Assume that he wishes to have an equal load on each foot. How is this shift in his centre of gravity accomplished? ANS: . whose mass is 80 kg. A 100 N force stretches a spring from 12 cm to 14 cm.11.11. p.262 42.5 LOC: EM1. In a Hooke’s Law experiment.Taking moments about M: REF: K/U KEY: FOP 6. What is the spring constant? ANS: REF: K/U MSC: P OBJ: 4. assuming that the spring remains within its elastic limit? ANS: . a force of 160 N produces a stretch of 8.08 KEY: FOP 6. What will be the length of the spring when a force of 500 N is applied. p.00 cm in a spring.262 MSC: P 41. 80 mm in diameter and 1.3 mm by a force of 18 N.08 KEY: FOP 6. A wire 0.262 43. supports a mass of 1500 kg. ANS: . p. How much is the bar stretched? ANS: REF: K/U KEY: FOP 6. A rectangular steel bar 5. REF: K/U MSC: P OBJ: 4.8 m long was stretched 1. Find Young’s Modulus for the wire. with a cross-sectional area of 1.The new length is 12 cm + 10 cm = 22 cm.11.5 LOC: EM1.00 m long.11.3 × 10–3 m2. p.262 MSC: P 44. If the force applied by the rower on the oar is 80 N.3 m from the oarlock. supports a load one-half the force of gravity on a 65 kg person.262 MSC: P 46. p. (a) What is the force of the oar on the water? (b) What is the force on the oarlock? . What is the compression experienced by the leg bone? ANS: REF: K/U KEY: FOP 6. The oar makes contact with the water at an average of 1. An oar is held by a rower at 0.0 cm 2. A 48 cm long leg bone.11.262 MSC: P 45. with a cross-sectional area of 5.REF: K/U KEY: FOP 6. p.35 m from the oarlock.11. answer the following. If the mass of the figure on the bottom left is 125 g.ANS: (a) Taking torques about P: (b) Taking components at right angles to the oar: REF: K/U KEY: FOP 6.258 MSC: P 47. A designer wishes to construct a mobile (as shown in the diagram) using strings and light rods (of negligible mass). p. what are the required masses of the other figures so that the rods are horizontal? ANS: .11. MC KEY: FOP 6. 150 N [W] (d) 6. (a) 160 N [east].0 N [N30°E].0 N [N45°W].REF: PROBLEM K/U. 10.0 N [W10°S] ANS: (a) 160 N [E] + 120 N [W] = 40 N [E] or 160 N [E] – 120 N [E] = 40 N [E] (b) (c) . find the additional force required to maintain static equilibrium. Forces as given below are acting on a common point.11. 160 N [north] (c) 100 N [N45°E]. p.259 MSC: P 1. 12. Using two different solutions for each combination of forces. 120 N [west] (b) 200 N [east]. (d) Using Components: . A helium balloon is attached to the middle of a light fishing-line anchored at both ends. ANS: .REF: K/U MSC: P OBJ: 2. If the upward force of the balloon is 0. p.2 LOC: FMV.1. find the tension in the fishing-line.02 KEY: FOP 6.208 2. as illustrated.05 N. as illustrated. The sign for Pierre’s restaurant has a mass of 82 kg. and a wire at 40° to the horizontal. ANS: Solution 1: Components Taking vertical components: Taking horizontal components: . Find the tension in the wire and the compression in the steel rod. which supports no weight.209 3. It is held out from the wall by a light horizontal steel rod. REF: K/U MSC: P OBJ: 2.Note that the tension in each string is about 6 times the balloon force.2 LOC: FMV. p.02 KEY: FOP 6.1. (b) the additional tension in the tie. held together by one single pin at each vertex. what is (a) the additional compression in each rafter. If a 100 kg mass is hung from the apex of the rafters.02 KEY: FOP 6.209 4. as illustrated. and (c) the additional compression in each stud? ANS: . A simple roof truss is composed of two rafters and a horizontal tie.2 and the compression in the rod is LOC: FMV. MC MSC: P OBJ: 2.Solution 2: Triangle of Forces Thus the tension in the wire is REF: K/U.1. p. KEY: FOP 6. MC MSC: P OBJ: 2. p. Alternate Solution: Total force down on both studs is 980 N. Thus force on each stud is REF: K/U.2 LOC: FMV.1. what is the maximum force in the string just before the box moves? ANS: . If the coefficient of static friction is 0.40.02 . (c) The additional compression at the stud will be the vertical components of the compressive force on the rafter at the stud.(a) Taking vertical components: (b) Tension in the tie will be the horizontal component of the compressive force on the rafter at the stud.209 5. A horizontal pull is being exerted on a 10 kg box with a string. REF: K/U MSC: SP OBJ: 2.0 kg block is placed on an adjustable ramp.40? ANS: Taking components parallel to the ramp and making up positive. Taking horizontal components Taking vertical components.02 KEY: FOP 6. . To what maximum angle can the ramp be raised before the block just begins to slide. if the coefficient of friction is 0. A 1.4 LOC: FMV.Since the box is in equilibrium.210 6. p.2. REF: K/U MSC: SP OBJ: 2.2. What is the maximum mass of the mouse. Kc is horizontal and fastened to the cheese.50 kg block of cheese sits on a level table. if the cheese and the mouse remain in equilibrium? ANS: .211 7.4 LOC: FMV. as shown.01 KEY: FOP 6. p.60. Three strings are tied together in a knot at K. and again making up positive. Therefore the ramp must be elevated 22° to the horizontal in order for the block to begin to slide down it. Thus. Km hangs vertically. supporting a mouse. Kw angles up to the wall at 30° to the horizontal. The coefficient of static friction is 0. A 0.Taking components normal to the ramp. this vector equation gives us Taking vertical components. with up as positive. with right as positive. we know that Taking horizontal components. gives us Since Tm = Fg.7 N.Since the knot is in equilibrium. and its mass would be . the force of gravity on the mouse is 1. A flat rock with a mass of 20 kg rests on a rough horizontal plane.01 KEY: FOP 6. p.4 LOC: FMV.211 8. if the force acts in the following ways.2.22.REF: K/U MSC: SP OBJ: 2. find the least force required to just move the rock. If the coefficient of friction is 0. (a) horizontally (b) at an angle of 37° above the horizontal (c) at an angle of 37° below the horizontal ANS: (a) (b) Taking vertical components: Thus Taking horizontal components: . p. If the combined mass of the seated child and the sleigh is 75 kg.01 KEY: FOP 6. to just get the sleigh to move with the third child seated in it.4 LOC: FMV. One child exerts a maximum force of 40 N and another 58 N. The angles the horizontal ropes make with the direction of travel are 30° and 20°. and the sleigh moves in a straight line. on ropes. Three children are playing in the snow with a sleigh.2.(c) Taking vertical components: Taking horizontal components: REF: K/U MSC: P OBJ: 2.212 9. respectively. what is the coefficient of friction between the sleigh and the snow? . A 2.212 10.01 KEY: FOP 6.ANS: REF: K/U MSC: P OBJ: 2. what is the coefficient of friction? (b) . p.4 LOC: FMV. as illustrated.0 kg wooden block is attached to a 0.50 kg provides the minimum force required to just get the block to move.2.50 kg mass by a string passing through a frictionless pulley. (a) If the mass of 0. what minimum mass. will just get the block moving? ANS: (a) (b) . attached to the string.If the same plane surface is now inclined 20° to the horizontal. will the block slide down the plane? ANS: .2.5 kg block rests on a plane inclined at 15° to the horizontal.Taking components along the plane with down the slope negative.213 11.4 LOC: FMV. p.02 KEY: FOP 6. If the coefficient of static friction for the surface is 0.30. A 2. The mass required is REF: K/U MSC: P OBJ: 2. REF: K/U MSC: P OBJ: 2. is supported by a hinge at A and a sloping rope at B.2.Since . 1. the block will not move down the slope.4 LOC: FMV.2 m long. ANS: Taking torques about A: .02 KEY: FOP 6. Find the tension in the rope and the vertical and horizontal reactions of the hinge if the rope is just about to open the trap door.213 12. p. A 41 kg trapdoor. Alternate Solution: Therefore 15° is less than angle of repose and object will not slide. 0 m long is supported by sawhorses A and B.219 MSC: P 13.0 m from B. A 60 kg crate is placed on the plank. as shown.3. 2. find the reaction forces at A and B. A 40 kg plank 6. (b) If the box and sawhorse B are moved as illustrated.Taking horizontal components: Taking vertical components: REF: K/U. (a) Find the reaction forces at A and B. ANS: (a) Taking torques about A: . MC KEY: FOP 6. p. 220 MSC: P 14. a rope is tied horizontally 0. is leaning against a wall.50 m from the bottom of the ladder to the wall. Because both the wall and the floor are frictionless. whose mass is 20 kg.0 m from the wall and at a slope of 60° to the floor. A 72 kg plasterer climbs threequarters of the way up the ladder and stops. A 12.3.Taking torques about B: Alternate Solution: (b) Taking torques about A: Taking vertical components: REF: K/U KEY: FOP 6. with its base 6.0 m ladder. p. What is the tension in the rope and what are the reactive forces at the wall and the floor? . ANS: (a) Taking moments about A: . 0 cm diameter sphere weighing 18 N.89 × 103 kg/m3). A cheerleader’s baton consists of a 4.0 cm and length 40 cm is turned down on a lathe to one-half of its radius for a distance of 30 cm from one end. REF: K/U KEY: FOP 6. Find the centre of gravity. ANS: . since the baton is balanced in equilibrium.Taking horizontal components: REF: K/U KEY: FOP 6.4.6 N and an 8. p.3.4 N and whose length is 86 cm.220 MSC: P 15. Calculate the centre of gravity. p. ANS: Applying Fnet = 0 in the vertical direction gives Taking A as the point of rotation. measured from the thicker end (density of steel = 7. joined by a thin tube whose weight is 4.0 cm diameter sphere weighing 3. Therefore The centre of gravity is located 71 cm from A.222 MSC: SP 16. A cylindrical steel rod of radius 2. opened in Toronto in 1976. to solid rock.18 × 108 kg.4. p. The foundation has threefold symmetry.224 MSC: P 17.0 m. Made of reinforced concrete. and pulls horizontally so that the tower pivots on two corners of the foundation. A massive foundation plunges an additional 16. do not consider the effects of the earth and rock around the foundation. and the centre of gravity is located about 75 m above the bottom of the foundation.7 m below the base. with the three corners lying on a circle of radius 55.2 m from the base of the tower to the highest antenna on its communication mast.Taking torques about P: REF: K/U KEY: FOP 6. what tension is needed to just start the tower tipping? Convert your answer to a fraction of the tower’s mass. or slides. The CN Tower. For the purposes of the problem given below. (a) A blimp fastens a cable to the uppermost tip of the tower. . The total tower mass is 1. breaks. but assume the tower and foundation to be resting on horizontal bedrock. is still the tallest free-standing structure in the world. So long as nothing bends. it soars 553. (b) Suppose the blimp succeeds in starting the tower tipping. This is a large force. but it is only about 5% of the force of gravity on the tower. which is located one-half the circle radius from the pivot line (see diagram). ANS: (a) The centre of gravity of the tower will be directly above the centre of the foundation. and carefully pulls it over to balance at the critical tipping angle: calculate this angle. . as seen below. 0 m steel guitar string has a diameter of 0. p. as illustrated. Thus. What is the value of the tension in the string if it stretches 0. the tower will fall back to its upright position if the tipping angle is less than 20°. A 1.5.20 cm.228 MSC: SP 18.(b) The critical angle will occur when the centre of gravity of the tower is vertically above the supporting edge. At the critical tipping angle. REF: K/U KEY: FOP 6.20 cm when tightened? ANS: . 236 MSC: SP 19.REF: K/U KEY: FOP 6. p.30 cm to cause its length to increase by 0.7. How much mass would have to be suspended on a vertical copper wire with a diameter of 0.20%? ANS: . The chandelier has a mass of 750 kg and is to be supported by a single steel cable. what is the minimum diameter of the cable (steel tensile strength = 5. p.236 MSC: SP 20. .7.0 × 108 N/m2)? ANS: For a safety factor of 5. Assuming a safety factor of 5.REF: K/U KEY: FOP 6. A lighting engineer is designing a large chandelier for the lobby of a concert hall. 5 m.236 MSC: SP 21.REF: K/U KEY: FOP 6.7. (a) What is the stress in the column? (b) What is the strain in the column? (c) If the column is 12 m high. A marble column supporting a load of 1.0 × 105 kg has a diameter of 0. how much will it be compressed? ANS: (a) . p. 0 m long is elongated 0.7.7. If a brass wire with an identical diameter were used to support the same load.(b) (c) REF: K/U KEY: FOP 6.238 MSC: P . p. REF: K/U KEY: FOP 6.238 MSC: P 22. An aluminum wire 1. what would its initial length have been for it to have had the same extension? ANS: But and stress is equal. p.02 cm when under stress. 5 × 104 N? ANS: REF: K/U KEY: FOP 6. A 50 cm long rod with a diameter of 1. How much will a 40 m steel elevator cable stretch if its diameter is 2. What is the Young’s Modulus of the rod? What is the material? ANS: .1 × 10–3 mm.0 cm and it is supporting a static load of 1.7. p.23.0 cm is used to support a load of 100 N along its length.238 MSC: P 24. As a result. the rod is compressed 9. calculate the diameter of the cable required if the maximum acceleration is 1. the substance is probably aluminum.0 m/s 2. REF: K/U KEY: FOP 6.7.238 MSC: P 25. Allowing for a safety factor of 4. p. .Thus. ANS: Tensile strength of steel is Allowing for a safety factor. A steel cable supports a 3000 kg elevator. (a) Find the tension in the biceps muscle. when the forearm is outstretched. p.7.238 MSC: P 26.REF: K/U KEY: FOP 6. as illustrated. (c) What is the reaction force of the radius on the humerus. (b) Find the tension in the biceps muscle if the forearm forms an angle of 45° with the horizontal. and its centre of gravity is located 14 cm from the joint. The biceps muscle is attached to the radius bone at 4. at the joint. The mass of the forearm is 2. and the mass is 32 cm from the joint.0 cm from the joint.0 kg. A 10 kg mass is held in a hand. as in (a)? ANS: (a) . The humerus bone pushes down on the forearm at the elbow. REF: K/U. that is. Take vertical components. MC KEY: FOP 6. (a) the tension in the biceps muscle (b) the reactive force of the radius on the humerus. It follows that the tension in the biceps muscle will be the same for all angles of the elbow. The result will be the same—a tension in the biceps muscle of 8.9. when the same mass is held. with up as positive. except that each factor will be multiplied by sin 45°.4 kg. Given the dimensions on the diagram of the forearm holding a mass of 5. since there is equilibrium. at the joint . the forces act at 45° to the forearm. p.(b) When the elbow forms an angle of 45°.244 MSC: SP 27. as illustrated.5 × 102 N. The torque equation in (a) will be the same. (c) The net force on the forearm is zero. find the following. MC KEY: FOP 6. REF: K/U.11.2 × 104 N [W]. while the ox pulls with a force of 1.6 × 104 N [W30°N]. What is the resultant force on the stump? ANS: Using Components: . The horse pulls with a force of 1.256 MSC: P 28. Thus Taking the vertical components. A pioneer farmer hitches a horse and an ox to a stump in order to remove it from a field. p.ANS: (a) Taking torques about P: (b) The net force on the forearm is zero. as illustrated.REF: K/U MSC: P OBJ: 2. A patient is in neck traction with skull calipers.20.256 29. given that mass of the head is 4.02 KEY: FOP 6. Find the maximum value of the suspended mass that will cause no tension or other force in the neck. p.2 kg and the coefficient of friction between the bed and the head is 0.11.2 LOC: FMV. ANS: Components of F: . ANS: . MC MSC: P OBJ: 2. p.Taking vertical components: Taking horizontal components: But.02 KEY: FOP 6. A 0.2 LOC: FMV.256 30. REF: K/U.60 g spider hangs on its thread from the branch of a tree. A horizontal wind blows the spider and the thread to an angle of 35° from the vertical.11. Find the force of the wind on the spider and the tension in the thread. To keep Robin from being captured.2 LOC: FMV. Assuming that Robin’s mass is 45.0 m rope hangs slack between hooks of equal height on adjacent buildings 13.0 kg and the rope withstands the initial impulse.02 KEY: FOP 6.Taking vertical components: Taking horizontal components: REF: K/U MSC: P OBJ: 2.0 m apart. p.257 31. knowing that a 17. what is the tension in each part of the rope when equilibrium is established? ANS: . Robin grabs the rope and hangs on at a point 5.11. Batman tosses him out of a third-storey window.0 m from one end. 0 m pole with a mass of 4. Assume that the centre of gravity of the pole is located at its midpoint. .02 KEY: FOP 6. (a) Find the vertical force of each hand on the pole when the pole is carried horizontally. A pole-vaulter runs down the track at constant velocity.257 32. (b) Find the force on the pole-vaulter’s upper hand during the high carry if it is perpendicular to the pole.11.0 m apart.Taking horizontal components: Taking vertical components: Substitute (1) in (2): REF: K/U MSC: P OBJ: 2.2 LOC: FMV. p. carrying a 6. and the vaulter’s hands are placed 1. as shown.0 kg. ANS: (a) Taking torques about P: Taking vertical components: (b) Taking torques about P: . and the masses are as follows: Mike 60 kg. have been lowered over the side of a ship on an 8.260 MSC: P 34.11. Find the tension in the two ropes.40. p. Two painters. The distance relationships are shown on the diagram. If the coefficient of static friction between the ladder and the ground is 0. Mike and Ned.258 MSC: P 33. ANS: Taking torques about P: Taking vertical components: REF: K/U KEY: FOP 6. p. Ned 80 kg.REF: K/U KEY: FOP 6.5 kg leans against a smooth wall at an angle of 50° to the horizontal. A 6. and the plank 40 kg. how far up the ladder can a 70 kg woman climb before the ladder starts to slip? ANS: .0 m ladder with a mass of 12.0 m plank supported by two ropes P and Q.11. 0 × 10–3 m in diameter is suspended from a solid support.11.Taking torques about P: REF: K/U KEY: FOP 6.0 m long and 2.261 MSC: P 35. p. A copper wire 2. A mass of 10 kg is hung from the bottom of the wire. What will be the stress and strain in the wire? ANS: . 02% of its length? ANS: . A brass rod has a diameter of 5. What force will stretch it by 0.0 mm.REF: K/U KEY: FOP 6. p.11.262 MSC: P 36. (a) Find the tensions in the cables. as shown in the diagram.11. (b) If both cables have a diameter of 2. p.262 MSC: P 37.0 mm. how much does each cable stretch? ANS: (a) Taking horizontal components: Taking vertical components: . A load of 250 kg is supported by two steel cables.REF: K/U KEY: FOP 6. 800 N [E15°S].262 MSC: P 38.6.232 39.08 KEY: FOP 6.REF: K/U. MC KEY: FOP 6.11.25 kg mass suspended from it. (a) 20 N [W] and 16 N [N30°W] (b) 1200 N [N65°E]. 1000 N [S30°W] ANS: (a) . A spring whose force constant is 48 N/m has a 0. Find the resultant of the following concurrent forces. p.5 LOC: EM1. p. What is the extension of the spring? ANS: If the spring obeys Hooke’s Law. REF: K/U MSC: SP OBJ: 4. 02 KEY: FOP 6.11.256 40. p. ANS: . Calculate the force at the hands and toes of a 58 kg athlete holding a push-up position.(b) Using Components: REF: K/U MSC: P OBJ: 2.2 LOC: FMV. p.0 m diving board with a mass of 30 kg is supported at X and Y. Calculate the forces at X and Y when a 50 kg diver stands at the outer end of the board. A 5. ANS: Taking torques about x: .259 MSC: P 41.Taking torques about P: Taking vertical components: REF: K/U KEY: FOP 6.11. Find the tension in the quadriceps tendon for the situation shown.11. ANS: Taking moments about P: Force on bottom of vertebrae: Horizontal components: . assuming that the weight of the leg is ignored. p.260 MSC: P 42.Taking vertical components: REF: K/U KEY: FOP 6. 9. KEY: FOP 6. p.249 MSC: P .Vertical components: Note the direction of REF: K/U is really below the horizontal.