Question

Jason is pulling a box across the room. He is pulling with a force of 24 newtons and his arm is making a 71 angle with the horizontal. If the box weighs 16 newtons what is the netforce on the box in the vertical direction? Treat up as the positive direction and down as the negative direction

Answer 1

We are given that a box is being pulled with a force of 24 Newtons as shown in the diagram above. We are asked to determine the net vertical force on the box. To do that we will first determine the vertical component of the box using the following triangle:

We can use the trigonometric function sine to determine the vertical component of the force, like this:

`\sin 72=(F_y)/(F)`

Multiplying both sides by "F":

`F\sin 72=F_y`

Replacing the value of F:

`24\sin 72=F_y`

Solving the operations:

`22.83N=F_y`

Now, we determine the sum of the forces acting in the vertical direction. The forces acting in the vertical direction are the weight in the negative direction and the vertical component of the force. Therefore, the sum of the forces is:

`\Sigma F_y=F_y-W`

Replacing the values we get:

`\Sigma F_y=22.83N-16N`

Solving the operations we get:

`\Sigma F_y=6.83N`

Therefore, the net vertical force is 6.83 Newtons.