A block with a weight of 3.0 Nis at rest on a horizontal surface. A 1.0 Nupward force is applied to the block by means of an attached vertical string. (a) What is the magnitude and (b) What is the direction of the force of the block on the horizontal surface?

• Weight of block, W=3.0 N .
• Upward force applied to the block, $F_{vertical}=1.0\mathrm{N}$.

The free-body diagram often referred to as FBD, represents all the forces acting on the body along with the directions. The diagram gives the general idea of net forces on the body.

Here, the vertical forces acting on the block are - the weight of the block, the pushing force of the horizontal surface to the block, and the pulling force of the vertical string. Using all these forces, we can find the required force.

Formula:

Since, the total forces acting on the body is balanced in the given direction,

$\underset{}{\overset{}{\sum F=0}}$ (i)

Free Body Diagram of the Block

(from equation (i) and the body diagram)

$$\begin{gathered} \left(\begin{array}{c}\mathrm{Vertical}\mathrm{force}\mathrm{by}\\ \mathrm{means}\mathrm{of}\mathrm{string}\end{array}\right)+\left(\begin{array}{c}\mathrm{Pushing}\mathrm{force}\mathrm{by}\mathrm{horizontal}\\ \mathrm{surface}\mathrm{on}\mathrm{block}\end{array}\right)-\left(\begin{array}{c}\mathrm{Weight}\mathrm{of}\\ \mathrm{block}\end{array}\right)=0 \\ \left(\begin{array}{c}\mathrm{Pushing}\mathrm{force}\mathrm{by}\mathrm{horizontal}\\ \mathrm{surface}\mathrm{on}\mathrm{block}\end{array}\right)=3.0\mathrm{N}-1.0\mathrm{N} \\ =2.0\mathrm{N} \end{gathered}$$

Using Newton’s third law, the force exerted by the block on the surface has the same magnitude so,

Force exerted by the block on the horizontal surface$=2.0\mathrm{N}$.

Hence, the value of force is$2.0\mathrm{N}$

From the above calculations and the free body diagram, it can be seen that the force exerted by the block on the horizontal surface is in the downward direction.