Question

Jason a box across the roomHe with a force of 19 newtons and his arm is making a 39 angle with the horizontalthe weighs 17 newtons the force on the box in the vertical direction Treat up as the positive directionand down as the negative direction

Answer 1

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

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

The weight of the box, W=17 N

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

The magnitude of the reaction force is equal to its weight. Let us call this force R.

Thus, R=17 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,

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

Therefore, the net force acting in the vertical direction is,

`F_n=R+F_v`

On substituting the known values in the above equation,

`F=17+11.96=28.96\text{ N}`

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