Torque question?

Question

A 6kg rock is suspended by a massless string from one end of a 7m measuring stick. What is the mass of the measuring stick if it is balanced by a support force at the 1m mark? Answer in kg.

Answer 1

assuming scale of uniform mass/length
let mass of scale = m kg
mass/length (mo) = m/7 (kg/m)
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taking moment of weight (torque) about the supported point
6m section >> mass (of 6m) = 6(m/7) kg will act at (6/2 = 3m)
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1m section >> mass (of 1m) = 1(m/7) kg will act at (1/2 = 0.5m)
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6kg mass will act at [1m]
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so
6(m/7)g *3meter = (m/7)g*(0.5meter) + 6g *(1meter)
18(m/7) = (m/14) + 6
m[36 - 1] /14 = 6
m = 2.4 kg

Answer 2

You have two unknowns here: the mass of the stick and the magnitude of the support force, which I assume is 1 m from the rock (rather than 1 m from the other end). You also have two equations: vertical force balance and moment balance. That means that you can solve for both.

Start by drawing a free body diagram. The weight of the rock is a downwards force, as is the unknown weight of the stick itself. Remember that a mass m has a weight equal to mg, where g is 9.8 m/s^2. The stick's weight acts at its center of mass, which is 3.5 m from each end. The support force is upwards, 1 m from the rock, between the rock and the center of mass of the stick. Call the mass of the stick M and the support force F.

Summing the forces in the vertical direction (upwards positive) and setting it equal to zero, you get F + M*9.8 - 6*9.8 = 0. Summing the moments about the support force (counterclockwise positive) and setting them equal to zero, you get 6*9.8*1 - F*2.5 = 0 (Note that the rock is 1 m from the support force, and the center of mass of the stick is 2.5 m from the support, on the other side.). You can use the second equation to solve for F, and then use the value for F in the first equation to solve for M, the mass of the stick.

Answer 3

Can't solve for the mass of the stick unless we know the support force at 1m.