A block of mass m = 1.40 kg is at rest on a ramp of mass M = 3.90 kg which, in turn, is at rest on a frictionless horizontal surface (see (a) in the figure). The block and the ramp are aligned so that each has its center of mass located at x = 0. When released, the block slides down the ramp to the left and the ramp, also free to slide on the frictionless surface, slides to the right (see (b) in the figure). Calculate xamo (in m), the distance the ramp has moved to the right, when xlock = -0.500 m.

A block of mass m = 1.40 kg is at rest on a ramp of mass M = 3.90 kg which, in turn, is at rest on a frictionless horizontal surface (see (a) in the figure). The block and the ramp are aligned so that each has its center of mass located at x = 0. When released, the block slides down the ramp to the left and the ramp, also free to slide on the frictionless surface, slides to the right (see (b) in the figure). Calculate xamo (in m), the distance the ramp has moved to the right, when xlock = -0.500 m.

International Edition---engineering Mechanics: Statics, 4th Edition

4th Edition

ISBN:9781305501607

Author:Andrew Pytel And Jaan Kiusalaas

Publisher:Andrew Pytel And Jaan Kiusalaas

Chapter7: Dry Friction

Section: Chapter Questions

Problem 7.78P: The figure shows a steel bar being processed by a rolling mill. Given that P=80kN and r =0.016,...

See similar textbooks

Similar questions

5- (Work, Energy) The block M in Fig (a) weighs 500N and has a downward velocity of 5 m/s at the instant the force P=1 kN is applied. The coefficient of friction between the arm AB and the rotating body is 0,4. The rotating body weighs 800 N and has a centroidal radius of gyration of 50 cm. Find the distance the weight travels before coming to rest. 30cm | 60cm 30ст B W M S0cm ||
A 50-kg smooth cylinder is at rest in a box that has smooth walls at right angles to each other. If the box is tiped up to a 45 degree angle, what is the reaction of the bottom of the box on the cylinder?
A board sits in equilibrium. On the left end, there is a wire that supports the board from the ceiling, and to the right, there is a sawhorse that supports the board from the ground. The sawhorse is a distance d= 1/5l from the right edge of the board. There is a block with mass ms= 4.5kg that is a distance 3/4l from the right edge of the board. Finally the board has a mass mb= 11kg. c) Write down Newton's 2nd law. Put the equation in terms of mb, ms, g, T, and Fn. Where T is the tension in the wire, and Fn is the normal force of the sawhorse on the board. d) Write down Newton's 2nd Law for rotations. Put the equation in terms of l, mb, ms, g, T, and Fn.
Penny is adjusting the position of a stand up piano of mass mp = 155 kg in her living room. The piano is lp = 1.35 m in length. The piano is currently at an angle of θp = 36 degrees to the wall. Penny wants to rotate the piano across the carpeted floor so that it is flat up against the wall. To move the piano, Penny pushes on it at the point furthest from the wall. This piano does not have wheels, so you can assume that the friction between the piano and the rug acts at the center of mass of the piano. Randomized Variables mp = 155 kglp = 1.35 mθp = 36 degrees a. Write an expression for the minimum magnitude of the force Fs in N Penny needs to exert on the piano to get it moving. Assume the corner of the piano on the wall doesn't slide and the static friction between the rug and the piano is μs. b. The coefficient of kinetic friction between the carpet and the piano is μk = 0.27. Once the piano starts moving, calculate the torque τp in N⋅m that Penny needs to apply to keep moving the…
A block of mass m = 2.00 kg is attached to a spring of force constant k = 555 N/m as shown in the figure below. The block is pulled to a position x, = 4.90 cm to the right of equilibrium and released from rest. x= 0 x= x; (a) Find the speed the block has as it passes through equilibrium if the horizontal surface is frictionless. m/s (b) Find the speed the block has as it passes through equilibrium (for the first time) if the coefficient of friction between block and surface is u, = 0.350. m/s
A 6-m long ladder has a mass of 13 kg., and its center of gravity is 2.4 m from the bottom along the ladder. The ladder is resting on a 5° upward sloping floor at A and on a wall inclined to the right at 5° from the vertical at B, so that it makes an angle of 60° with the horizontal. How far up along the ladder can a 78-kg man climb before the ladder is on the verge of slipping? The angle of friction at all contact surfaces is 15°. Also determine the total reaction at A and B in Newton. Note: Point A is at the left of point B.
A block of mass m = 2.00 kg is attached to a spring of force constant k = 525 N/m as shown in the figure below. The block is pulled to a position x = 5.80 cm to the right of equilibrium and released from rest. k x=0 x=x; (a) Find the speed the block has as it passes through equilibrium if the horizontal surface is frictionless. m/s (b) Find the speed the block has as it passes through equilibrium (for the first time) if the coefficient of friction between block and surface is μk m/s Need Help? Read It Watch It Submit Answer = 0.350.
In order to burgle a local McDonald's, the Hamburglar must climb onto the roof. To do this, a ladder of length 3.31 m and mass 21.2 kg is placed leaning against a vertical wall, as shown in the image below. There is negligible friction between the ladder and the wall, but the same is not true for the ladder and the ground: us ≠ 0 between those two surfaces. Even with that friction, the ladder will start slipping if the angle ø ladder makes relative to the ground is less than 71.0°. Assuming the ladder is uniform (so that its center of gravity is located at its geometric center), what must be the maximum static frictional force (in N) that exists between the bottom of the ladder and the ground? Explain how you solved the problem involving a ladder leaning against a wall. Be sure to state what your known and unknown quantities are, what concepts were applied, and what equations were used!
A 30kg mass is tied to a uniform rod whose mass and length are 80-kg and 8-m, as shown below. The rod is attached to the floor via a hinge and makes an angle of 70° relative to the ground; a massless string is also attached to the rod at the ¼ point of the rod and makes an angle of 20° relative to the horizontal. What tension must be provided by the rope so that the system remains in stationary? If at some point the string breaks, what will be the angular acceleration of the system the instant that the string breaks?
3- (WORK AND ENRGY) The block M in Fig (a) weighs 500N and has a downward velocity of 5 m/s at the instant the force P-1 kN is applied. The coefficient of friction between the arm AB and the rotating body is 0,4. The rotating body weighs 800 N and has a centroidal radius of gyration of 50 cm. Find the distance the weight travels before coming to rest. 30cm. 60cm 30cm
An ideal cord (no mass and no elongation under load) is wrapped around a disc of radius r (r=50 cm, J=0,4 kgm²). The free end of the cord is connected to the mass of m=3kg. The disc is pin jointed and no friction is arisen at the pin. The cord does not slide on the disc and the system is at rest initially. When the system is released the mass, due to the gravity, starts a to move with acceleration a. Find the acceleration of the mass, and the change in the linear momentum of the mass as well as in the angular momentum of the disc during 3 seconds. g=9.81 m/s²
A 6-m long ladder has a mass of 10 kg, and its center of gravity is 2.4 m from the bottom along the ladder. The ladder is resting on a 5° upward sloping floor at A and on a wall inclined to the right at 5° from the vertical at B, so that it makes an angle of 60° with the horizontal. How far up along the ladder can a 78-kg man climb before the ladder is on the verge of slipping? The angle of friction at all contact surfaces is 15°. Also determine the total reaction at A and B in Newton. Note: Point A is at the left of point B. SOLVE ASAP