Question

A body of weight $50\text{N}$ is dragged slowly with constant velocity on a horizontal surface as shown in the figure, with a force of $F=28.2\text{N}$ at an angle of ${45}^{\circ }$ with horizontal. The net contact force acting on the body will be

• A. $10\sqrt{17}\text{N}$
• B. $10\sqrt{19}\text{N}$
• C. $10\sqrt{13}\text{N}$
• D. $10\sqrt{12}\text{N}$

Answer 1

The correct option is C $10\sqrt{13}\text{N}$

From the free body diagram of the block


It is given in the problem that the body moves with constant velocity hence acceleration, $a=0$ an so the frictional force will be equal to the force applied in horizontal direction
$f=F\cos {45}^{\text{o}}=\frac{28.2}{\sqrt{2}}=20\text{N}$ where $(\sqrt{2}\approx 1.41)$

Now for calculating for the normal force, we balance vertical forces
$N+F\sin {45}^{\text{o}}=50\Rightarrow N=50-20=30\text{N}$

Hence the net contact force acting on the body will be
$\sqrt{f^2+N^2}=\sqrt{{20}^2+{30}^2}=10\sqrt{13}\text{N}$