M D P L sin x FN L sin C W L sin C fl cos D 0.

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eglect friction between the ladder and the wall. α Solution: he weight of the ladder is D 18 g D 176.58. he weight of the person is D 90

1 789 roblem 9.26 he masses of the ladder and person are 18 kg and 90 kg, respectively. he center of mass of the 4-m ladder is at its midpoint. If D 30, what is the minimum coefficient of static friction between the ladder and the floor necessary for the person to climb to the top of the ladder? eglect friction between the ladder and the wall. α Solution: he weight of the ladder is D 18 g D he weight of the person is D 90 g D Leth be the distance along the ladder of the person s center of mass, and L be the length of the ladder. he horizontal distance is. he sum of the moments about the top of the ladder: D L sin L sin C L sin C fl cos D 0. 2 α rom the sum of the forces, Y D D 0, from which the normal force at the foot of the ladder is D C. Substitute, solve for the friction force, and reduce algebraically: ( h f D L C 1 ) 2 tan. t the top of the ladder, h D 1, hence L ( f D C ) tan D 883 C D R t impending slip, f D s D s C, from which s D f C D D α f

2 825 roblem 9.76 Suppose that in roblem 9.75, weighs 800 lb and weighs 400 lb. he coefficients of friction between all of the contacting surfaces are s D 0.15 and k D ill remain in place if the force is removed? Solution: he equilibrium conditions are: or the bo : Denote the normal force eerted by the wall by, and the normal force eerted by the wedge by. he friction forces oppose motion. y D Ccos C s sin C s D 0, DC s cos sin C D 0. ( comparison with the equilibrium conditions for roblem 9.75 will show that the friction forces are reversed, since for slippage the bo will move downward, and the wedge to the right.) he strategy is to solve these equations for the required s to keep the wedge in place when D 0. he solution D 0, D lb, D lb and s D Since the value of s required to hold the wedge in place is greater than the value given, the wedge will slip out. or the wedge. Denote the normal force on the lower surface by. D s cos s cos C sin C sin D 0. y D cos C cos s sin C s sin w D 0. µ S µ S µ S µ S roblem 9.77 etween and, s D 0.20, and between and C, s D etween C and the wall, s D he weights D 20 lb and C D 80 lb. hat force is required to start C moving upward? Solution: he active contact surfaces are between the wall and C, between the wedge and C, and between the wedge and. or the weight C : Denote the normal force eerted by the wall by, andthe normal force between and C by. Denote the several coefficients of static friction by subscripts. he equilibrium conditions are: C y D C C C D 0, 15 D C C D 0. or the wedge : Denote the normal force between and by. y D C cos sin D 0. D C cos sin D 0. C hese four equations in four unknowns are solved: D 15.2 lb, 15 D 84.6 lb, D lb, and D 66.9 lb C µ s µ s µ s µ s

3 824 roblem 9.74 Suppose that between all contacting surfaces in roblem 9.73, s D 0.32 and k D eglect the weights of the 5 wedges. If a force D 800 is required to move to the right at a constant rate, what is the mass of? Solution: he free body diagrams of the left wedge and the combined right wedge and crate are as shown. he equilibrium equations are edge: D cos 5 C 0.3 sin 5 D 0, D 1180, D 1150, D 3820, and m D 343 kg. y D 0.3 C sin 5 C 0.3 cos 5 D 0, edge and bo: D cos sin D 0, m (9.81) y D sin cos m D 0. Solving them, we obtain roblem 9.75 he bo has a mass of 80 kg, and the wedge has a mass of 40 kg. etween all contacting surfaces, s D 0.15 and k D hat force is required to raise at a constant rate? Solution: rom the free-body diagrams shown, the equilibrium equations are o : sin k cos D 0, cos k sin k D 0. edge : sin C k cos C sin C k cos D 0 cos k sin cos C k sin w D 0. Solving with D , w D , and k D 0.12, we obtain D 845, µ k µ k µ k µ k D 247, D 1252, and D 612.

791 roblem 9.29 he disk weighs 50 lb. eglect the weight of the bar. he coefficients of friction between the disk and the floor are s D 0.6 and k D 0.4.

hat couple is necessary to rotate the disk at a constant rate?

D 8 100 20 50 C 20 D 0, 100 lb 8 in 12 in 5 in from which D 90 lb. he friction force is f D s. he moment eerted by the friction force is D s R D 0.

(b) he moment required to rotate the disk at a constant rate is 8 in 12 in 100 lb 5 in K D k R D 0.4 5 90 D 180 in lb f roblem 9.30 he cylinder has weight.

hat is the largest couple that can be applied to the stationary cylinder without causing it to rotate? R Solution: ssume impending slip.

4 791 roblem 9.29 he disk weighs 50 lb. eglect the weight of the bar. he coefficients of friction between the disk and the floor are s D 0.6 and k D 0.4. (a) (b) hat is the largest couple that can be applied to the stationary disk without causing it to start rotating? hat couple is necessary to rotate the disk at a constant rate? 8 in 12 in 100 lb 5 in Solution: he normal force at the point of contact is found from the sum of moments about the pin support. D C 20 D 0, 100 lb 8 in 12 in 5 in from which D 90 lb. he friction force is f D s. he moment eerted by the friction force is D s R D D 270 in lb his is the moment to be overcome at impending slip. (b) he moment required to rotate the disk at a constant rate is 8 in 12 in 100 lb 5 in K D k R D D 180 in lb f roblem 9.30 he cylinder has weight. he coefficient of static friction between the cylinder and the floor and between the cylinder and the wall is s. hat is the largest couple that can be applied to the stationary cylinder without causing it to rotate? R Solution: ssume impending slip. he force opposing rotation is the sum of the friction force at the wall and at the floor. Denote the normal force at the wall by and the normal force on the floor by.romthesumofforces: y D s C D 0, and D s D 0. R Solve these two simultaneous equations to obtain: D 1 C 2, s and D s 1 C 2. s he sum of moments about the center of the cylinder is µ s C D app s R s R D 0. Substitute and solve: ( ) 1 C s app D s R 1 C 2. s µ s t impending slip, this is the maimum moment that can be applied to the cylinder.

5 857 roblem he 20-kg bo is held in equilibrium on the inclined surface by the force acting on the rope wrapped over the fied cylinder. he coefficient of static friction between the bo and the inclined surface is 0.1. he coefficient of static friction between the rope and the cylinder is Determine the largest value of that will not cause the bo to slip up the inclined surface. 20 Solution: ssuming that slip of the bo up the surface impends. he free body diagrams of the bo and rope around the cylinder are as shown. rom the equilibrium equations D cos sin cos20 D 0, y D sin C cos sin D 0 20 we obtain D Equation (9.17) is D e s where s D 0.05 and ˇ D / rad. Solving for we obtain D 101. y 20 (20)(9.81) roblem In roblem 9.131, determine the smallest value of necessary to hold the bo in equilibrium on the inclined surface. Solution: In this case, we assume that slip of the bo down the surface impends. his requires reversing the direction of the friction force in the free body diagram of roblem he friction now acts up the surface and the friction on the drum is reversed. See the free body diagrams. rom the equilibrium equations, y 20 X D cos sin 20 C 0.1 cos 20 D 0, (20)(9.81) 0.1 and Y D sin C cos 20 C 0.1 sin D 0. Solving, we obtain D e can now use this to find the force that must be applied to the rope to keep the bo from slipping down the plane. Eq. (9.17) is D e sˇ, where s D 0.05 and ˇ D / rad. Solving for, we obtain D

6 860 roblem he spring eerts a 320- force on the left pulley. he coefficient of static friction between the flat belt and the pulleys is s D 0.5. he right pulley cannot rotate. hat is the largest couple that can be eerted on the left pulley without causing the belt to slip? 100 mm 40 mm 260 mm Solution: he angle of the belt relative to the horizontal is ( ) D sin 1 D radians mm 40 mm or the right pulley the angle of contact is ˇright D 2 D radians. he sum of the horizontal components of the tensions equals the force eerted by the spring: 260 mm D upper C lower cos D 320. Since the angle of contact is less on the right pulley, it should slip there first. t impending slip, the tensions are related by upper D lower e sˇright D lower. Substitute and solve: lower 1 C e D 320 cos, from which lower D 68.34, and upper D he moment applied to the wheel on the right is applied D R upper lower D D m roblem Suppose that the belt in roblem is a V-belt with angle D 30. hat is the largest couple that can be eerted on the left pulley without causing the belt to slip? Solution: Use the results of the solution to roblem 9.136, as applicable: he angle of contact with the right pulley is ˇright D radians. he sum of the horizontal components of the tensions equals the force eerted by the spring: D upper C lower cos D 320. Since the angle of contact is less on the right pulley, it will slip there. t impending slip, the tensions are related by upper D lower e 9,where for brevity 9 D sˇright ( ) D sin 2 has been substituted. Substitute and solve: lower 1 C e 9 D 320 cos, from which lower D 1.86, and upper D he moment applied to the wheel on the right is applied D R upper lower D D m

STATICS. Friction VECTOR MECHANICS FOR ENGINEERS: Eighth Edition CHAPTER. Ferdinand P. Beer E. Russell Johnston, Jr.

STATICS. Friction VECTOR MECHANICS FOR ENGINEERS: Eighth Edition CHAPTER. Ferdinand P. Beer E. Russell Johnston, Jr. Eighth E 8 Friction CHAPTER VECTOR MECHANICS FOR ENGINEERS: STATICS Ferdinand P. Beer E. Russell Johnston, Jr. Lecture Notes: J. Walt Oler Texas Tech University Contents Introduction Laws of Dry Friction.