Question

A crowbar 2m long is pivoted about a point 0.1m from its tip . Calculate the least force which must be applied to its other end to displace the load of 100 kgf.

Answer 1

Answer:

I don't know the answer

Answer 2

Answer:

1900 kgf

Explanation:

Given,

Load arm = 0.1 m

Effort Arm = 2 - 0.1 = 1.9 m

Now,

V.R = (Effort Arm/Load Arm)

     = 1.9/0.1

    = 19

Thus,

M.A = 19

(Load/Effort) = 19

Given, Effort = 100 kg.

⇒ (L/100) = 19

⇒ L = 1900 kgf

Therefore,

Least Force = 1900 kgf

Hope it helps!