Question

In the figure, suppose the length L of the uniform bar is 3.2 m and its weight is 220 N. Also, let the block's weight W = 270 N and the angle θ = 45˚. The wire can withstand a maximum tension of 450 N. (a) What is the maximum possible distance x before the wire breaks? With the block placed at this maximum x, what are the (b) horizontal and (c) vertical components of the force on the bar from the hinge at A?

Answer 1

Answer:

a)   x = 2.46 m

b)   318.2 N

c)    177.8 N

Explanation:

Need to resolve the tension of the string at end say B.

The vertical upward force at B due to tension is 450 sin 45°.  

Using Principle of Moments, with the pivot at A,

Anti clockwise moments = Clockwise moments

450 sin 45° X 3.2   = 220 X (3.2/2) + (270 X x)  

x = 2.46 m

(b) The horizontal force is only due to the wire's tension, so it is  

450 cos 45° = 318.2 N

(c) total downward forces = 270 + 220 = 496 N  

Total upward forces = 450 sin 45° (at B) + upForce (at A)

Equating, upForce = 496 - 318.2  

= 177.8 N