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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? Group of answer choices 4o 8o 11o 18o

Sagot :

Answer:

The correct choice is 8o.

Explanation:

The given weight is hanging at the mid point of the rope and the buildings are at the same level, Obtain the given equation by equating the vertical components of force,

2Tsin(φ)=W

where φ is the angle made by rope with the horizontal.

Given T=1800 N and W=500 N, the value of φ is calculated as follows,

2*1800*sin(φ)=500

sin(φ)=0.1388

φ=7.90

φ≈8o

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