Question

Find the force diagram, the box is moving along the horizontal direction.where F_N is the normal force

Answer 1

Given data:

* The value of the force F_1 acting at 20 degree is,

$F_1=60\text{ N}$

* The value of the force F_2 is,

$F_2=10\text{ N}$

* The mass of the box is 8 kg.

Solution:

As box is not moving along the vertical direction, thus, the normal force acting on the body must balancing the y-comonents of the forces acting on the body,

The equation for the normal force is,

$F_N+F_1\sin (20^{\circ}_{})-mg=0$

Here the negative sign is indicating the direction force towards downward, and the positive sign indicates the direction of force towards the upward direction,

Substituting the known values,

$\begin{gathered} F_N+60\times\sin (20^{\circ})-(8\times9.8)=0 \\ F_N+20.52-78.4=0 \\ F_N-57.88=0 \\ F_N=57.88\text{ N} \end{gathered}$

Thus, the normal force acting on the box is 57.88 N.