Question

1. (9 points) an apple crate of mass 23 kg accelerates along a surface as the crate is pulled with a force of 14.5 n as shown in the drawing. there is a friction force of 7.0 n. (a) show a complete free - body diagram with all forces acting on the box. (b) what is the weight of the crate? (c) what is the x - component (horizontal) of the 14.5 n force? (you will need trigonometry to answer this question!!)

Explanation:

(a) Brief Explanations:

• Gravitational Force (\(F_g\)): Acts vertically downward. Formula: \(F_g = mg\) (where \(m\) is mass and \(g = 9.8\ m/s^{2}\)).
• Normal Force (\(F_N\)): Acts vertically upward, perpendicular to the surface.
• Applied Force (\(F_{app}\)): Has a magnitude of \(14.5\ N\) and is at an angle of \(19^{\circ}\) above the horizontal.
• Frictional Force (\(F_f\)): Acts horizontally opposite to the direction of motion (opposite to the horizontal component of the applied force).

(b) Step - by - Step Format:

Step1: Recall the formula for weight

The weight \(W\) of an object is given by \(W=mg\), where \(m = 23\ kg\) and \(g = 9.8\ m/s^{2}\)

Step2: Calculate the weight

Substitute the values into the formula: \(W=(23\ kg)\times(9.8\ m/s^{2})\)
\(W = 225.4\ N\)

(c) Step - by - Step Format:

Step1: Recall the formula for the x - component of a force

The x - component of a force \(F\) at an angle \(\theta\) is given by \(F_x=F\cos\theta\). Here, \(F = 14.5\ N\) and \(\theta=19^{\circ}\)

Step2: Calculate the x - component

\(F_x=(14.5\ N)\cos(19^{\circ})\)
Using a calculator, \(\cos(19^{\circ})\approx0.946\)
\(F_x=(14.5\ N)\times0.946 = 13.727\ N\)

Answer:

• (a) Draw four forces as described above (gravitational, normal, applied, frictional) on the free - body diagram.
• (b) \(225.4\ N\)
• (c) \(13.727\ N\)