Question

A block weighing 5.0 N is at rest on a horizontal surface. A 1.0 N upward force is applied to the block by an attached helium filled balloon with a massless string. What are the magnitude and direction of the force of the horizontal surface on the block?

Also, what would the free body diagram look like?

Answer 1

The surface provides an upward normal of 4.0 N. See Explanation for description of the free body diagram.

Explanation:

The block continues to be at rest on the horizontal surface. Therefore all forces must sum to zero, yielding `F_"net" = 0`.

There are 3 forces acting on the block:

1. Its weight is a downward 5.0 N force
2. The balloon lifts with a force of 1 N
3. The surface provides an upward normal force, `F_N` that we have to calculate.

Let the upward direction be the positive direction.

`F_"net" = -5.0 N + 1 N + F_N = 0`
`F_N = 5.0 N - 1.0 N = 4.0 N`

The free body diagram would show upward forces of 1 N & 4 N and a downward force of 5 N.

I hope this helps,
Steve