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Sampson swings the 1,200N weight about a vertical axis in a horizontal circle. The length of the...

Question:

Sampson swings the 1,200N weight about a vertical axis in a horizontal circle. The length of the attached rope is 1.25 m. What is the resultant force acting on the weight?

Equilibrium of Forces :

In the equilibrium of forces condition, the net force acting on the object is equal to zero and the object either remains in the rest condition or the object must be moving in the uniform motion.

In the equilibrium condition, the sum of the forces acting in the horizontal direction and vertical direction is equal to zero.

Answer and Explanation:

Given

Weight of the object is {eq}W = 1200\ N {/eq}

length of the rope is {eq}l = 1.25\ m {/eq}

The tension in the string can be calculated as:

{eq}T=\frac{mg}{\cos\theta}\\ T=\frac{1200}{\cos 40^\circ}\\ T=1566.48\ N {/eq}

The centripetal force acting on the block must be the net force acting on the block:

{eq}F=T\sin\theta\\ F=1566.48\sin 40^\circ\\ F\approx 1007\ N {/eq}

Thus, the resultant force acting on the weight is 1007 N


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