Answer to Question #213391 in Mechanics | Relativity for Dell

Question

A 350N, uniform, 1.50m bar is suspended horizontally by two vertical cables at each end. Cable A can support a maximum tension of 500.0 N without breaking, and cable B can support up to 400.0 N. You want to place a small weight on this bar.

(a) What is the heaviest weight you can put on without breaking either cable, and

(b) Where should you put this weight?

Answer 1

Rigid object of equilibrium:-

A rigid object in equilibrium exhibit no translation for angular acceleration the net external force acting on it is zero and the net external torque on it is zero about any axis$\sum F_{ext}=0$

$\sum \tau_{max}=0$

The first condition is that condition for translation equilibrium and second is the condition for rotation equilibrium

Gives

$W_{bar}=350N$

I=1.50m

$T_A=500N,T_B=400N$

Let upward be positive apply the first condition for translation equilibrium from equation (1) the bar in the vertical direction

$$\sum F_y=T_A+T_B-350N-W=0$$

$$W=T_A+T_B-350=500+400-350=550N$$

Part (b) let counterclockwise rotation be positive apply the second condition for rotational equilibrium from equation to about an Axis through point A

$$\sum \tau_{ext}=T_B(l-x)-350(\frac{l}{2}-x)-T_Ax=0$$

$x=\frac{T_Bl-(350)\times\frac{l}{2}}{T_B+350+T_A}$

Put value

$$x=\frac{400\times1.50-(350)\frac{1.50}{2}}{400+350+500}=0.614m$$