Question

Jason is pulling a box across the room. He is pulling with force of 19 newtons and is making a 39 angle with the horizontalthe box 17 newtons what the net force on the box in the vertical direction Treat up as the positive direction down as the negative direction.

Answer 1

The force with which Jason is pulling the box, F=19 N

The angle at Jason is applying the force, θ=39°

The weight of the box, W=17 N

There are two forces acting in the vertical direction. One, the vertical component of the force applied by Jason. Two, the reaction force applied by the ground on the box due to the weight of the box.

Let the reaction force be R.

Then the magnitude of the reaction force will be,

`R=17\text{ N}`

The vertical component of the force applied by Jason is given by,

`F_v=F\sin \theta`

On substituting the known values in the above equation,

`F_v=19*\sin 39^(\circ)=11.96\text{ N}`

Therefore the net force acting in the vertical direction is,

`\begin{gathered} F_n=R+F_v \\ =17+11.96 \\ =28.96\text{ N} \end{gathered}`

Therefore the net force acting in the vertical direction is 28.96 N