Find the weight of the lightest block D that can be used to

Find the weight of the lightest block D that can be used to support the 200-lb uniform pole ABC in the position shown.

Solution

Let the weight of the block D =m

Normal Force Fn = m x g

µ = Coefficient of friction= 0.2                                                            

A- Frictional Force Ff= µ x m x g= 0.2 x m x g                           

To Find Frictional force at point B

Sin = 8/10= 0.8 = 53

Distance AB= Sq Root ( 6 ² + 10 ²) = 10                   

Weight of Pole ABC= 200 lb                                  

Weight at point B= 200 x 10/15= 133.3                   

B - Frictional force at B= µ x M X g X cos (90- )= 0.2 x 133.3 x g x 0.8

Equating A=B the value of m= 106.66 lb

The Weight of block D should be 106.66 lb