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A 500 N weight is hung at the middle of a rope attached to two buildings at the same level. If the breaks in the tension exceed 1800 N, what is the minimum angle the rope can make with the horizontal?

Sagot :

LammettHashidn

Not sure what you mean by "breaks in the tension" but I suspect you mean the rope will come apart if the tension in the rope exceeds 1800 N.

In the free body diagram for the 500 N weight, we have a figure Y with the net force equations

• horizontal net force:

∑ F[hor] = T₁ cos(θ) - T₂ cos(θ) = 0

• vertical net force:

∑ F[ver] = T₁ sin(θ) + T₂ sin(θ) - 500 N = 0

From the first equation, it follows that T₁ = T₂, so I'll denote their magnitude by T alone. From the second equation, we have

2 T sin(θ) = 500 N

and if the maximum permissible tension is T = 1800 N, it follows that

sin(θ) = (500 N) / (3600 N)   ⇒   θ = arcsin(5/36) ≈ 7.9°

is the smallest angle the rope can make with the horizontal.

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