Force required. A uniform ladder is 10m long and weighs 200 N. The ladder leans

Question

Force required.

A uniform ladder is 10m long and weighs 200 N. The ladder leans against a ,vertical, frictionless wall at a point 8 m above the ground. A horizontal force F is applied to the ladder 2m from the base of the ladder (as measured along the ladder), If the applied force. F, is 50N, what is the force of the ground on the ladder?(b) If the coefficient of static friction between the ladder and the ground is 0.38, what minimum value of F will cause the ladder to just to start to move toward the wall?

Explanation / Answer

sum torque about the bottom
sin theta = 8/10
torque = 2 *50* sin theta - F*10*sin theta - 200*5*cos(theta) = 0

2*50*8/10 -F*10*8/10 + 200*5*sqrt(1-(8/10)^2)=0
F=85

sum forces in the x
F bottom +50-85 = 0
F bottom = 35

b) when it just starts to move friction = uN

sum forces in the y:
N - 200 = 0
N = 200
so friction = 0.38*200=76 N
so sum forces in the x:
F - Ftop -76 = 0
F top = F - 76
sum torques about bottom
2*F*8/10 -(F - 76)*10*8/10 + 200*5*sqrt(1-(8/10)^2)=0
F=188.75 N

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