Question

A 250-lb block is subjected to a horizontal force P. The coefficient of friction between the block and surface is µs = 0.2. Determine the force P required to start moving the block up the incline

Answer 1

Answer:

force required to push the block = 219.714 lb

Explanation:

GIVEN DATA:

weight W of block = 250 lb

coefficient of friction = 0.2

consider equilibrium condition in x direction

$P*cos(30)-W*sin(30)-\mu _{s}N = 0$

$P*0.866-0.2N = 125$.........................(1)

consider equilibrium condition in Y direction

$N-Wcos(30)-Psin(30)= 0$

$N-0.5P=216.503$.....................(2)

SOLVING 1 and 2 equation we get N value

N = 326.36 lb

putting N value in either equation we get force required to push the block = 219.714 lb