414 When 50 g lifts 100 g

When the weight was hung onto a vertical string, the tension in the string was only equal to the weight.

However, when the weight was hung on a horizontal string, the horizontal string cannot remain horizontal. This is because a horizontal string cannot provide any upward component of tension to balance the downward pull of the weight.


At equilibrium, the string makes an angle θ with the vertical.

Since the vertical forces at point X should balance, 2Tcosθ = mg.

Do you see that the tension in the string is actually larger than the weight?

At point Y, we know that T = Mg. so we can work out that 2Mgcosθ = mg, which leads to the result that cosθ = m/(2M). In this video where m/M is 0.5, θ works out to be 75°.

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