Question

A body of mass $600g$ slides on a rough horizontal surface if the frictional force is $2N$. Find the magnitude of contact force, where g is $10m{s}^{-2}$.

Answer 1

Step 1: Given data

1. Mass of the body, m = $600g$.
2. The frictional force exerted on the body, N = $2N$.
3. Acceleration due to gravity, g = $10m{s}^{-2}$.

Step 2: Contact force

1. When two bodies are in contact with each other, the force exerted on the bodies is called contact force.
2. The frictional force is one type of contact force.
3. The contact force due to friction is, $F=\sqrt{f^2+N^2}$, where f is the frictional force and N is the normal force on the body.
4. And the value of normal force is the same as weight but directed opposite to the weight.
5. Normal force, N = mg, m is the mass of the body and g is the acceleration due to gravity.

Step 3: Finding the magnitude of contact force

The normal force on the body is $N=0.6\times 10=6$ Newton.

Let F be the contact force on the body, directed along with R as shown in the figure.

Now, the contact force is

$$\begin{gathered} N=\sqrt{\left(2^2+6^2\right)}=\sqrt{\left(4+36\right)} \\ \Rightarrow N=\sqrt{40}=6.32N \end{gathered}$$

Step 4: Diagram

The magnitude of contact force on the body is $6.32N$.