Question

A cubical block is subjected to three forces $F_1=20\text{N}$ acting at an angle of ${45}^{\circ }$ with horizontal and lying in the diagonal plane of the cube, $F_2=30\text{N}$ along $y-$axis and $F_3=40\text{N}$ acting in the negative $x-$direction as shown. Find the magnitude of friction force acting on the body.

• A. $10\sqrt{29}\text{N}$
• B. $10\sqrt{27}\text{N}$
• C. $10\sqrt{30}\text{N}$
• D. $50\text{N}$

Answer 1

The correct option is D $50\text{N}$

Given, $F_1=20N$, $F_2=30N$ and $F_3=40N$
Resolving the forces along the axes we get,
$F_1$ has three components:
(i) Along the vertical $=F_1\cos {45}^{\circ }=\frac{F_1}{\sqrt{2}}$
(ii) Along $F_2=\left(F_1\sin {45}^{\circ }\right)\cos {45}^{\circ }=\frac{F_1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=\frac{F_1}{2}$
(iii) Along $F_3=\left(F_1\cos {45}^{\circ }\right)\cos {45}^{\circ }=\frac{F_1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=\frac{F_1}{2}$
Hence, the net horizontal force acting in the x-y plane on the block is
$\sqrt{{(F_2+\frac{F_1}{2})}^2+{(F_3-\frac{F_1}{2})}^2}$
$=\sqrt{(30+10{)}^2+(40-10{)}^2}=50\text{N}$
Hence, frictional force will act opposite to this force.
Thus, the frictional force is $f=50N$