So, during physics lab we were talking about torque and solving problems (this i

Question

So, during physics lab we were talking about torque and solving problems (this isnt like a typical physics lab, its quite torture-some) We were balancing washers and paperclips on a scale. We called the unit of these washers and paperclips together as one washerclip (one of each) This is the question:

Suppose you had an object that you suspected had a mass of one half of a washerclip. Describe how you could use the balance and a single washerclip to check this. I assume it has to do with weighing it and dividing, but i need to phrase it the right way.

How could you use the balance to check that an object had a mass of one-fifth of one of your washerclips. ------ Again, with weighing and dividing..or maybe multiplying. My math is kind of off today, sorry. Please help me phrase this a way the TA will like. Thankssss!

Explanation / Answer

Let m be the mass of the washerclip and is placed at one of the end of the scale Let L be the length of the scale If the object of mass m/2 is placed at a distance x from point of pivot, for the balance, the point of pivot should be selected sothat the net torque is zero Therefore (m/2)x = m(L-x) x/2 = L-x 3x/2 = L x = 2L/3 If the paperclip balances the object with pivot at 2L/3 from the end, then we can say that the object has half the mass of the paper clip
If the object of mass m/5 is placed at a distance x from point of pivot, for the balance, the point of pivot should be selected sothat the net torque is zero Therefore (m/5)x = m(L-x) x/5 = L-x 6x/5 = L x = 5L/6 If the paperclip balances the object with pivot at 5L/6 from the end, then we can say that the object has half the mass of the paper clip
If the paperclip balances the object with pivot at 2L/3 from the end, then we can say that the object has half the mass of the paper clip
If the object of mass m/5 is placed at a distance x from point of pivot, for the balance, the point of pivot should be selected sothat the net torque is zero Therefore (m/5)x = m(L-x) x/5 = L-x 6x/5 = L x = 5L/6 If the paperclip balances the object with pivot at 5L/6 from the end, then we can say that the object has half the mass of the paper clip
If the object of mass m/5 is placed at a distance x from point of pivot, for the balance, the point of pivot should be selected sothat the net torque is zero Therefore (m/5)x = m(L-x) x/5 = L-x 6x/5 = L x = 5L/6 If the paperclip balances the object with pivot at 5L/6 from the end, then we can say that the object has half the mass of the paper clip
If the object of mass m/5 is placed at a distance x from point of pivot, for the balance, the point of pivot should be selected sothat the net torque is zero Therefore (m/5)x = m(L-x) x/5 = L-x 6x/5 = L x = 5L/6 If the paperclip balances the object with pivot at 5L/6 from the end, then we can say that the object has half the mass of the paper clip If the paperclip balances the object with pivot at 5L/6 from the end, then we can say that the object has half the mass of the paper clip

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