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

force required to push the block = 219.714 lb

Step-by-step 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