No problem! Let's continue with the solution.
We know that the center of mass (C) for a uniform mass measuring stick is in the middle. Since the measuring stick is 6m long, the center of mass C is at 3m. Now, we can use the moments equation:
M1 x L1 = M2 x L2
where:
M1 = 6 kg (mass of rock)
L1 = 1 m (distance from support to rock)
M2 = mass of measuring stick (which we want to find)
L2 = (C - 1 m) = (3 m - 1 m) = 2 m (distance from support to center of mass of measuring stick)
Now we can solve for M2:
6 kg * 1 m = M2 * 2 m
6 kg = M2 * 2 kg/m
M2 = 6 kg / 2 kg/m = 3 kg
So the mass of the measuring stick is 3 kg.
A 6 kg rock is suspended by a massless string from one end of a 6m measuring stick. What is the mass of the measuring stick if it is balanced by a support force at the 1m mark? answer in units of kg. Thanks for any help
Rather than just posting the question, you should post your suggested answer or thoughts as well.
Assume that measuring stick is of umiform mass where is the centre of mass? (C)
Use moments
M1 x L1 = M2 x L2
M1=6 kg
L1=1 m
L2= (i.e C - 1 m)
find M2
I'll keep that it mind. Thank you very much!
1 answer
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