Question

The crate, which sets on a 10 degree incline, has a weight of 3433.5 N and is subjected to a towing force P acting at a 20 degree angle with the horizontal. If the coefficient of static friction is µs = 0.5, determine the force P needed to just start the crate moving down the plane.

Answer 1

Answer:

P = 981 N

Explanation:

Given

Angle of the incline θ = 10°

Angle of the towing force φ =20°

Weight of the crate W = 3433.5 N

The coefficient of static friction µ = 0.5

Solution

Forces Acting along the ramp

$\mu N = Pcos(10^{o}+10^{o})+Wsin10^{0} \\\mu N = Pcos30^{o}+Wsin10^{0}$

Forces acting perpendicular to the ramp

$N + Psin30^{o}=Wcos10^{0}\\\\N=Wcos10^{0}-Psin30^{o}$

Substituting the value of N we get

$\mu[Wcos10^{0}-Psin30^{o}] =Pcos30^{o}+Wsin10^{0}\\\\\mu Wcos10^{0}- \mu Psin30^{o} =Pcos30^{o}+Wsin10^{0}\\\\\mu Wcos10^{0}- Wsin10^{0}=Pcos30^{o}+\mu Psin30^{o} \\\\P=W\frac{\mu cos10^{0}- sin10^{0}}{cos30^{o}+\mu sin30^{o}} \\\\P=3433.5\frac{0.5 \times cos10^{0}- sin10^{0}}{cos30^{o}+0.5 \times sin30^{o}}\\\\P=980.66\\\\ P = 981 N$