Asked by carolyn

Consider the following system with a 2 kg mass sitting on a table, connected to both 1 kg and 3 kg masses by strings which go over two separate pulleys and are pulling on the 2 kg mass in opposite directions. What static friction coefficient would be required between the 2 kg mass and the table to stop the masses from accelerating? If the masses are in motion, what kinetic friction coefficient would be required between the 2 kg mass and the table to ensure the masses move at a constant velocity? Find the tension in both pieces of string for the case of zero friction
Answered by bobpursley
net force=2g-1g-2g*mu=0
solve for muStatic
same equation for moving, only mu is now muKinetic

Tension in both strings=weight of the mass hanging over.
Answered by bobpursley
net force=3g-1g-2g*mu=0
solve for muStatic
same equation for moving, only mu is now muKinetic

Tension in both strings=weight of the mass hanging over.

See correction above, I typed in the hanging mass as 2 and 1, when they should be 3kg and 1kg
Answered by dd
dsvfsv
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