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A small branch is wedged under a 200 kg rock and rests on a smaller object. The small object that creates a pivot point is called the fulcrum. The smaller object is 2.0 m from the large rock and the branch is 12.0 m long. If the mass of the branch is negligible, what force must be exerted on the free end to just barely lift the rock

2 Answers

3 votes

Answer:

F = 326.7 N

Step-by-step explanation:

User Gherkins
by
4.7k points
6 votes

Answer:

F = 326.7 N

Step-by-step explanation:

given data

mass m = 200 kg

distance d = 2 m

length L = 12 m

solution

we know force exerted by the weight of the rock that is

W = m × g ..............1

W = 200 × 9.8

W = 1960 N

and

equilibrium the sum of the moment about that is

∑Mf = F(cos∅) L - W (cos∅) d = 0

here ∅ is very small so cos∅ L = L and cos∅ d = D

so F × L - W × d = 0 .................2

put here value

F × 12 - 1960 × 2 = 0

solve it we get

F = 326.7 N

User EldadT
by
5.8k points