Question

A 350.0 N, uniform, 1.50 m bar is suspended horizontally by two vertical cables at each end. Cable A can support a maximum tension of 550.0 N without breaking, and cable B can support up to 300.0 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

Answer:

a. 500 N

b. x = 0.375

Explanation:

Given that

Weight of the bar is W = 350 N,

Length of the bar is L = 1.50 m

Tension in the string A is T1 = 550 N,

Tension in the string B is T2 = 300 N

a. The computation of heaviest weight is shown below:-

Here we will apply equilibrium conditions where Fy = 0

T1 + T2 -W' - W = 0

W' = 550 + 300 - 350

= 500 N

b. The computation of weight is shown below:-

we will assume x be the distance between A and mass

now, we will apply a torque about the point A

$= T1\times (x) - W\times (\frac{L}{2} - x) + T2\times (L-x) = 0$

So,

$x = \frac{( T2\times L - W\times \frac{L}{2} )}{( T1 - W + T2)}$

$x = \frac{( 300\times 1.50 - 350\times \frac{1.50}{2} )}{( 550 - 350 + 300)}$

x = 0.375

Therefore we have applied the above formula.