Question

End A of the bar AB in the fig rests on a frictionless horizontal surface, and end B is hinged. A horizontal force $\overrightarrow{\mathbf{F}}$of magnitude 220 N is exerted on end A. Ignore the weight of the bar. What are the horizontal and vertical components of the force exerted by the bar on the hinge at B?

Short answer

Horizontal component is 220 N to the right and vertical component is 293 N upward.

Step-by-step solution

The given data

Given that end A of the bar AB rests on a frictionless horizontal surface, and end B is hinged.

A horizontal force $\overrightarrow{F}$of magnitude 220 N is exerted on end A.

So horizontal force$F=220\text{N}$

Weight of the bar is negligible.

Formula used

Torque$\mathit{\tau }\mathbf{=}\mathit{F}\mathit{l}$

Where Fis force exerted and l is distance.

Find the horizontal and vertical component of force.

Let $H_{v}and{H}_h$are vertical and horizontal components of force exerted on the bar by the hinge.

Let nis the normal force exerted on the bar by the surface.

The components of force of the bar on the hinge will be equal in magnitude but opposite in direction.

Force of bar on hinge in horizontal direction $F_x=H_h=220\text{N}$

Force of bar on hinge in vertical direction$F_y=n-H_v$

Now, applying the condition of equilibrium,

Net torque is zero.

This implies

$$\begin{gathered} F\left(4\text{m}\right)-n\left(3\text{m}\right)=0 \\ \left(400\text{N}\right)\left(4\text{m}\right)-n\left(3\text{m}\right)=0 \\ n=293\text{N} \end{gathered}$$

Thus$H_v=293\text{N}$

Conclusion

Hence, horizontal component of force exerted on the bar by the hinge is 220 N to the right and vertical component is 293 N upward.