b)If the maximum force is applied and the block begins to slide, what is the mag

Question

b)If the maximum force is applied and the block begins to slide, what is the magnitude and direction of the force of friction now?

c) If the maximum force is applied and the block begins to slide, what is the acceleration of the block assuming that the tensional force for (a) is maintained? (This is related to b.)

d) After the block is sliding, I want to reduce the magnitude of the tensional force so that the block slides at a constant velocity. What is the tensional force required so that the block slides at a constant velocity?

Problem 9. Consider a 10 kg block at rest on a flat table with a string applying a tensional force directed 30 degrees above the x-axis as shown below. The coefficient of static friction between the block and the floor is 0.5 and the coefficient of kinetic friction between the block and the floor is 0.25. (This problem is similar to the example done in the video. At the end of the video it asks for you to find the force required to move the block at a constant velocity. If you tried it you should have determined that the force required was, 25N. If you did not figure this out I would suggest trying that before proceeding with this question.) 10 kg Floor

Explanation / Answer

a) Normal reaction on the block, N = mg - (T sin30o) = 10 * 9.8 - (T * 0.5) = 98 - T/2

Maximum static friction, fs = 0.5 * (98 - T/2) = 49 - T/4

Horizontal pulling force = T cos30o = 0.866T

So, pulling force is equal to maximum static friction just before sliding.

=> 0.866T = 49 - T/4

=> T = 43.9 N

b) Friction = 0.25 * (98 - 43.9/2) = 19 N (directed opposite to the movement of block)

c) Horizontal pulling force = 0.866T = 0.866 * 43.9 = 38 N

Kinetic friction = 19 N

So, acceleration, a = (38 - 19)/10 = 1.9 m/s2

d) For constant velocity, horizontal pulling force should be equal to kinetic friction.

Hence, 0.866T = 19

=> T = 21.94 N

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