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Expert-verified Found in: Page 117 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # A block with a weight of 3.0 N is at rest on a horizontal surface. A 1.0 N upward 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?

(a) 2.0 N is the magnitude of the force exerted by the block on the horizontal surface

(b) The direction of the force exerted by the block on the horizontal surface is downwards.

See the step by step solution

## Step 1: The given data

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

## Step 2: Understanding the concept of force in the free body diagram

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)

## Step 3: (a) Calculation of the magnitude of the force

Free Body Diagram of the Block (from equation (i) and the body diagram)

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

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 $2.0\mathrm{N}$

## Step 4: (b) Finding the direction of the normal force

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. ### Want to see more solutions like these? 