Question

a force \\( \vec{p} \\) pulls on a crate of mass \\( m \\) that is in contact with a rough surface. the figure shows the magnitudes and directions of the forces that act on the crate in this situation. \\( \vec{w} \\) represents the weight of the crate. \\( \vec{f}_n \\) represents the normal force on the crate, and \\( \vec{f} \\) represents the frictional force. which statement best describes the motion of the crate?\\( \vec{f}_n \\) \\( \vec{p} = 160 \\, \text{n} \\) \\( \vec{f} = 80 \\, \text{n} \\) \\( 60^\circ \\) \\( \vec{w} = 196 \\, \text{n} \\)\
\\( \bigcirc \\) the crate must be moving with constant acceleration.\
\\( \bigcirc \\) the crate must be at rest.\
\\( \bigcirc \\) the crate may be either at rest or moving with constant velocity.\
\\( \bigcirc \\) the crate must be moving with constant velocity.

Explanation:

Step1: Analyze Horizontal Forces

First, find the horizontal component of force \(\vec{P}\). The formula for the horizontal component of a force \(P\) at an angle \(\theta\) is \(P_x = P\cos\theta\). Here, \(P = 160\,\text{N}\) and \(\theta = 60^\circ\), so \(P_x = 160\cos60^\circ\). Since \(\cos60^\circ = 0.5\), \(P_x = 160\times0.5 = 80\,\text{N}\). The frictional force \(f = 80\,\text{N}\) acts in the opposite direction. So the net horizontal force \(F_{\text{net},x}=P_x - f = 80 - 80 = 0\,\text{N}\).

Step2: Analyze Vertical Forces

The vertical component of \(\vec{P}\) is \(P_y = P\sin\theta = 160\sin60^\circ\approx160\times0.866 = 138.56\,\text{N}\). The weight \(W = 196\,\text{N}\) acts downward, and the normal force \(F_N\) acts upward. The net vertical force \(F_{\text{net},y}=F_N + P_y - W\). But for the purpose of motion, we care about net force. Since net horizontal force is zero and we assume vertical forces are balanced (as there's no acceleration mentioned vertically), the net force on the crate is zero.

Step3: Relate Net Force to Motion

According to Newton's first law, if the net force on an object is zero, the object will either be at rest or moving with a constant velocity (uniform motion). So the crate may be either at rest or moving with constant velocity.

Answer:

The crate may be either at rest or moving with constant velocity.