Question

a 2.5kg block initially at rest on horizontal surface. A horizontal force F of magnitude 6N and a vertical force P are the applied to the block. The coefficients of friction for the block and surface are Ms=0.4 and Mk=0.25.Determine the magnitude of the frictional force acting on the block if force P is a) 8N b) 10N c) 12N

Answer 1

(a) If P = 8N, the maximum static friction force is Ms*(M*g + P) =
(0.4)(24.5 + 8) = 13 N
Since the applied force is less than that, no motion will take place and the actual friction force will equal the applied horizontal force, 6N .

Answer 2

can you explain more

Answer 3

In each case, there are two vertical forces: the weight and the applied force P. Multiply the sum of them by 0.4 (the static friction coefficient) to get the maximum force required to overcome friction. If this force is not exceeded by F, the block will not move. In that case, the actual friction force equals F, to prevent motion.

Answer 4

mg = 9.8 m/s^2 * 2.5kg = 24.5N
F(applied) = 6.0N
F(normal) = 24.5N - P
F(static) = Ms*(24.5 - P)
If F(static) > F(applied) = 6.0N --> F(friction) = F(applied)
If F(applied) = 6.0N > F(static) --> F(kinetic) = Mk*(24.5 - P)
a) F(normal) = 24.5N - 8.0N = 16.5N
F(static) = 0.4*16.5N = 6.6N > F(applied) = 6.0N, so F(friction) = 6.0N
b) F(normal) = 24.5N - 10.0N = 14.5N
F(static) = 0.4*14.5N = 5.8N < F(applied), so...
F(friction) = Mk*14.5 = 3.625N
c) F(normal) = 24.5N - 12.0N = 12.5N
F(static) = 0.4*12.5N = 5.0N < F(applied), so...
F(friction) = Mk*12.5 = 3.125N