91影视

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

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

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

Weight of the bar is negligible.

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

Where Fis force exerted and l is distance.

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$H_v=293\text{N}$

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.