A uniform bar weighing 60N has a length of 4m and it is supported at its ends by a string. A load W=20N is placed at some point on the bar such that it is in equilibrium when the tension in the string at the left end is 80N,
a) locate tge position of W.
b) find the tension in the string at the right end.
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To solve this problem, we can use the principle of moments, which states that the sum of the clockwise moments about any point is equal to the sum of the anticlockwise moments about the same point, when the object is in equilibrium.
a) To locate the position of W, we need to find the distance of W from the left end of the bar. Let's assume this distance is x meters.
According to the principle of moments, we can set up an equation as follows:
Clockwise moment = Anticlockwise moment
(80 N) * (0 m) = (20 N) * (4 m - x)
Simplifying the equation: 0 = 80 N - 20 N * (4 m - x)
Expanding: 0 = 80 N - 80 N + 20 N * x
Further simplifying: 0 = 20 N * x
Since the weight of W is 20 N, and the distance from the left end is x, we can conclude that x = 0.
Therefore, W is located at the very left end of the bar.
b) To find the tension in the string at the right end, we can consider the whole bar as a system in equilibrium. The sum of the vertical forces has to be zero.
Let's consider the tension at the right end as T R.
Vertical force up = Vertical force down
T L + T R = Weight of the bar + Weight of W
T R = Weight of the bar + Weight of W - T L
T R = 60 N + 20 N - 80 N
T R = 0 N
Therefore, the tension in the string at the right end is zero.