Question

4.3 Calculate magnitude of force F. I have already made Freebody diagram

Answer 1

Answer:

1) Free Body Diagram is drawn below

2) Newton's first law of motion is stated below

3) Force, F = 182.2 N

Normal Force = 427.7 N

Frictional force = 171.08 N

Explanation:

The Free Body diagram is drawn below:

Newton's first law of motion states that a body will remain in a state of rest or uniform motion unless it is acted upon by an external source

The magnitude of the force F is calculated below

Since the object is at constant velocity, the acceleration, a = 0 m/s^2

$\begin{gathered} \sum F=ma \\ \\ Fcos20-F_r=50(0) \\ \\ Fcos20-F_k=0.........................* \\ \\ Note: \\ F_k=\mu(mg-Fsin20) \\ \\ F_k=0.4[(50*9.8)-0.34F] \\ \\ F_k=196-0.136F..................** \\ \\ Substitute\text{ \lparen **\rparen into \lparen *\rparen} \\ \\ Fcos20-(196-0.136F)=0 \\ \\ 0.94F-196+0.136F=0 \\ \\ 1.076F=196 \\ \\ F=\frac{196}{1.076} \\ \\ F=182.2N \\ \end{gathered}$

The normal force is calculated below

$\begin{gathered} \sum F_y=0 \\ \\ Fsin20-mg+F_N=0 \\ \\ 182.2sin20-50(9.8)+F_N=0 \\ \\ 62.3-490+F_N=0 \\ \\ F_N=490-62.3 \\ \\ F_N=427.7N \end{gathered}$

The frictional force is calculated below

$\begin{gathered} F_r=\mu_kF_N \\ \\ F_r=0.4(427.7) \\ \\ F_r=171.08N \end{gathered}$