Answer to Question #164170 in Mechanics | Relativity for 12-STEM

Question #164170

A uniform 80.0 N ladder 4.0 m long is placed against a frictionless wall with its base situated 2.0 from the wall. Find the forces exerted by the wall and by the ground on the ladder.


1
Expert's answer
2021-02-19T09:43:43-0500

Let normal reaction at endpoints be N1,N2N_1, N_2

Force of friction acting between ground and wall is f.

The weight of the ladder will act at the mid-point of the ladder.

Let angle made by the wall with ground is θ,cosθ=24=12\theta, cos\theta = \frac{2}{4}= \frac{1}{2}


Then for equilibrium in the horizontal and vertical direction

f=N2f = N_2 (1)

N1=mg=80NN_1 = mg = 80N (2)


Taking torque about point on ground,

N2lsinθ=mgl2cosθ    N2=23.1NN_2 lsin\theta = mg\frac{l}{2} cos\theta \implies N_2 = 23.1N





Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment