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

Question

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

bobpursley · Answer

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.

bobpursley · Answer

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

dd · Answer

dsvfsv