can anyone help me and show me how to do this physics?

Question

if R1 is 518.95 cm and R2 is 1200 cm, what is the force F is required to lift a mass of 60 kg at constant speed?

Answer 1

clarify R1 and R2; are u talking about universal law of gravitational force? (F = m1m2/r^2

Answer 2

Unfortunately, you failed to specify how the pulleys (R1 and R2) are interconnected with each other and the weight. However, making the assumption that you are winding the smaller pulley R1 and the weight is attached to the larger pulley R2, the mechanical advantage A = R2/R1 ~ 1200/500.

So your force (F) with the mechanical advantage (A) can lift FA = mg ~ 600 Newtons; so that F = 600(500/1200) ~ 25 Newtons to raise the 600 Newton weight.

(I rounded off things to make the arithmetic easier. You can plug in the real numbers if you need the precision.)

Answer 3

your question is slightly confusing. The answer is contained in the second part of the question...

"...what is the force, F, required to lift a mass of 60 kg at constant speed?"

The key clue here is the "constant speed". I am assuming that since you use the verb "to lift" then the only component of movement is vertically up in a straight line?

Since there is no acceleration of the 60kg mass (constant speed, remember) then the only force required to keep it moving vertically at a constant speed will be:

F = - mg

where m = 60 kg and g ~ 9.8 metres per second squared. g is called the "acceleration due to gravity".

If we take the direction of g (which is a vector) to be positive then F acts in the opposite direction so it is negative (note the minus sign above).

So...

F = - 60 kg times 9.8 metres per second squared.

I'm sure you can work out the rest.

I don't know what the first part of your question, R1 and R2 is all about. Maybe you need to provide some more detail. Is ther some kind of circular motion involved?