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ncrowley1818
Jan18-05, 06:05 PM
I have a homework problem, that is giving me some trouble.
I have to move two blocks that weigh (1.2kg), connected by a rope. They give static friction coefficients for each block. I know that I must overcome the static frictions to make them move. I think I need to add the coefficients and then do something with the weight of the blocks?? Any comments or help???

Thanks,
Nick

Doc Al
Jan18-05, 06:15 PM
Try considering the two blocks as a single system. What's the total friction force on the system just before it begins to move?

ncrowley1818
Jan18-05, 06:30 PM
I can add the static forces together .55+.4=.95 total static friction

Doc Al
Jan18-05, 06:40 PM
I cannot comment on your answer since you did not provide the coefficients of static friction. Also: Be sure to use proper units.

ncrowley1818
Jan18-05, 06:46 PM
the values are the coefficients of static friction on the two blocks, I added them together to get the total static friction of the total system. I know I must over come the total to make it move. F=ma???

Doc Al
Jan18-05, 06:55 PM
the values are the coefficients of static friction on the two blocks, I added them together to get the total static friction of the total system.
Are you saying that each block has a mass of 1.2 Kg and the coefficients of static friction are 0.55 and 0.45?

If so, find the frictional force on each block. What's the equation relating frictional force to the coefficient of friction?

ncrowley1818
Jan18-05, 07:03 PM
dont you take the

force of static friction/mass of object=coefficient static friction

Doc Al
Jan18-05, 07:17 PM
dont you take the

force of static friction/mass of object=coefficient static friction
No. The maximum force of static friction between two surfaces is given by:
f = \mu N, where N is the normal force pressing the surfaces together. For a block on a horizontal surface the normal force equals the weight of the block (mg), so f = \mu mg.