Question

A 50.0-kg box is being pulled along a horizontal surface by means of a rope that exerts a horizontal force of 250N. The coefficients of friction between the bottom of the machine and the floor are µk= 0.3 and µs= 0.4.a. Draw the free-body diagram for the box.b. ) Write the Newton's second law equations for the box below.Fx =Fy =c. Calculate the normal force acting on the box.d. ) Calculate the acceleration of the box.

Answer 1

(a). The free body diagram for the box is represented as,

Where F' is the frictional force, N is the normal reaction, and mg is the weight of the box.

(b). The normal force acting on the box is,

`\begin{gathered} N=mg \\ N=50*9.8 \\ N=490\text{ N} \end{gathered}`

The friction force acting on the box is,

`\begin{gathered} F^(\prime)=\mu_sN \\ F^(\prime)=0.4*490 \\ F^(\prime)=196\text{ N} \end{gathered}`

The forces acting along the x-axis is,

`\begin{gathered} F_x=F-F^(\prime) \\ F_x=250-196 \\ F_x=54\text{ N} \end{gathered}`

Thus, the force acting along the x-axis is 54 N.

The force acting along the y-axis is,

`\begin{gathered} F_y=N-mg \\ F_y=mg-mg \\ F_y=0\text{ N} \end{gathered}`

Thus, the force acting along the y-axis is zero.

(c). The normal force acting on the box is,

`N=490\text{ N}`

(d). According to the Newton's second law,

`F=ma`

For the force along the x-axis,

`F_x=ma`

Substituting the known values,

`\begin{gathered} 54=50* a \\ a=(54)/(50) \\ a=1.08ms^(-2) \end{gathered}`

Thus, the acceleration of the box is 1.08 meter per second squared along the x-axis.