Question

Determine the normal reaction in the given figure.
     

Answer 1

Hint: In the given figure, a force is acting on the block at an angle of ${{45}^{0}}$ which is a ‘pulling’ force. Now, since this force is at a certain angle with the horizontal, there exists a vertical component of this force. So, the net sum of the normal reaction due to the floor on the block and the vertical component of external force will balance the force of gravity, which is acting in downward direction.

Complete answer:
Let us assign some important terms that we are going to use in our problem.
Let the mass of the block be denoted by M. Then, its value has been given to us as:
$\Rightarrow M=5kg$
Let the external force be denoted by F. Its value is given as:
$\Rightarrow F=20N$
And, let the angle made by this external force with the horizontal be $\theta $ . Then, it is equal to:
$\Rightarrow \theta ={{45}^{0}}$
Now, we know that Normal reaction on the block will be in a vertically upwards direction, and since there is no motion of the block along the vertical direction, we need to balance all the forces in the vertical direction. Mathematically, this could be written as:
Assuming the normal reaction by the ground to be ${{N}_{R}}$ , we have:
$\Rightarrow {{N}_{R}}+F\sin \theta -mg=0$
Putting all the respective values in the above equation, we get:
$\begin{align}
  & \Rightarrow {{N}_{R}}=mg-F\sin \theta \\
 & \Rightarrow {{N}_{R}}=5\times 9.8-20\sin {{45}^{0}} \\
 & \Rightarrow {{N}_{R}}=49-\dfrac{20}{\sqrt{2}} \\
 & \Rightarrow {{N}_{R}}=49-\dfrac{20}{1.414} \\
 & \Rightarrow {{N}_{R}}=34.856N \\
\end{align}$
Hence, the net normal reaction by the ground on the block comes out to be 34.856N.

Note:
It is very important to know that Normal reaction always acts along the perpendicular of two surfaces in contact. Here, we were given the values of static and dynamic coefficients of friction to confuse us that Normal Reaction had something to do with friction. But that isn’t the case. It doesn’t matter if the block is stationary or moving, the normal reaction will stay the same.