3 blocks are connected by a massless inextensible cord over massless frictionless pulleys. On top of a table (in the middle) is a 51kg block. One mass hangs on the left (102kg) and another hangs on the right (M). The coefficient of static friction between the 51kg block and the table top is 0.62 and the system is in equilibrium (the 51kg block is in the middle of the table top, and the left & right blocks are hanging down across from each other). acceleration of gravity is 9.8 m/s^2.
Find the maximum value of M for which the system remains in equilibrium. answer in units of kg.
2006-09-30 14:22:12 · 2 answers · asked by Dee 4 in Science & Mathematics ➔ Physics
Tried to draw a picture w/ this but couldn't...Hope somebody can help.
2006-09-30 14:37:20 · update #1
The middle block can resist a force of 0.62*(51 kg)*(9.8 m/s^2) = 309.9 Newtons. The right and left blocks will have to have an unbalance such that 309.9 N are exerted on the middle block. W = 309.9 N = M*9.8. So M = 31.62 kg. If the right weight is 31.63 kg more, or less, than the left one, the system should start moving.
2006-09-30 15:30:10 · answer #1 · answered by sojsail 7 · 0⤊ 0⤋
Uhhh...what??? :)
2006-09-30 21:30:28 · answer #2 · answered by Anonymous · 0⤊ 0⤋