Answered

Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Our Q&A platform provides quick and trustworthy answers to your questions from experienced professionals in different areas of expertise. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

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 :

LammettHash

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.

Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.