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Korolek [52]
4 years ago
8

A 250-lb block is subjected to a horizontal force P. The coefficient of friction between the block and surface is µs = 0.2. Dete

rmine the force P required to start moving the block up the incline

Physics
1 answer:
aev [14]4 years ago
6 0

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

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