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marta [7]
4 years ago
3

A horizontal uniform plank is supported by ropes I and II at points P and Q, respectively, as shown above. The two ropes have ne

gligible mass. The tension in rope I is 150 N. The point at which rope II is attached to the plank is now moved to point R halfway between point Q and point C, the center of the plank. The plank remains horizontal. What are the new tensions in the two ropes?
The answer is T1=100N and T2=200N but I don't know the steps to solve this one. An explanation would be much appreciated.

Physics
1 answer:
umka21 [38]4 years ago
8 0

Explanation:

There are three forces on the plank.  T₁ pulling up at point P, T₂ pulling up at point Q, and W pulling down at point C.

Let's say the length of the plank is L.

Sum of forces in the y direction before rope II is moved:

∑F = ma

150 N + 150 N − W = 0

W = 300 N

Sum of moments about point P after rope II is moved:

∑τ = Iα

(T₁) (0) − (300 N) (L/2) + (T₂) (3L/4) = 0

-(300 N) (L/2) + (T₂) (3L/4) = 0

-(300 N) (1/2) + (T₂) (3/4) = 0

-150 N + 3/4 T₂ = 0

T₂ = 200 N

Sum of forces in the y direction:

∑F = ma

T₁ + 200 N − 300 N = 0

T₁ = 100 N

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