Question

problem 5: a pail of water hangs from a line right in the middle. the line is attached to two poles as shown in the diagram. the weight of the pail is w = 250 n, and the line sags by an angle of θ = 1.2°. (a) draw a free body diagram of the bucket. (b) write down the equations one gets from newton’s second law. (c) now find the tension (t) on the line. (d) if the sagging angle increases, what happens to the force of tension?

Explanation:

Step1: Draw free - body diagram

The bucket has its weight \(W\) acting downwards and two tension forces \(T\) acting at an angle \(\theta\) to the horizontal on either side of the bucket.

Step2: Apply Newton's second law in vertical direction

Since the bucket is in equilibrium (\(a = 0\)), \(\sum F_y=0\). The vertical components of the two tension forces balance the weight of the bucket. The vertical component of each tension force is \(T\sin\theta\). So, \(2T\sin\theta - W=0\).

Step3: Solve for tension \(T\)

From \(2T\sin\theta - W = 0\), we can re - arrange to get \(T=\frac{W}{2\sin\theta}\). Substituting \(W = 250\ N\) and \(\theta=1.2^{\circ}\), we have \(\sin\theta=\sin(1.2^{\circ})\approx0.0209\). Then \(T=\frac{250}{2\times0.0209}\approx5981\ N\).

Step4: Analyze effect of angle change on tension

From \(T=\frac{W}{2\sin\theta}\), as \(\theta\) increases, \(\sin\theta\) increases. Since \(W\) is constant, the value of \(T\) decreases.

Answer:

(a) A free - body diagram shows the weight \(W\) acting downwards and two tension forces \(T\) acting at an angle \(\theta\) to the horizontal on either side of the bucket.
(b) \(2T\sin\theta - W = 0\) (in vertical direction, as the bucket is in equilibrium).
(c) \(T\approx5981\ N\)
(d) The force of tension decreases.