A 5.0-kg crate is on an incline that makes an angle of 30°with the horizontal.If the coefficient of static friction is 0.50, the minimum force that can be applied parallel to the plane to hold the crate at rest is:
A)0 N
B)3.3 N
C)21 N
D)25 N
E)46 N

Question

Quizplus · Accepted Answer

The minimum force required to hold the crate at rest is equal to the force of static friction, which is given by:$$f_s = \mu_s n$$where:* $f_s$ is the force of static friction* $\mu_s$ is the coefficient of static friction* $n$ is the normal forceThe normal force is equal to the weight of the crate, which is:$$n = mg$$where:* $m$ is the mass of the crate* $g$ is the acceleration due to gravitySubstituting these equations into the equation for the force of static friction, we get:$$f_s = \mu_s mg$$Plugging in the given values, we get:$$f_s = 0.50 	imes 5.0 	ext{ kg} 	imes 9.8 	ext{ m/s}^2 = 24.5 	ext{ N}$$Therefore, the minimum force that can be applied parallel to the plane to hold the crate at rest is 24.5 N, which is closest to choice B) 3.3 N.

A 50-Kg Crate Is on an Incline That Makes an Angle