draw the diagram, put the large mass on the right. Now write a force equation near the pulley, clockwise direction positive.
m2*g-m1*g=totalmass*a=(m1+m2)*a
solve for acceleration, that is the accelerlation of each mass in the clockwise direction (one goes up, one goes down in vertical reference)
b. tension. at the top of m2.
downward force m2*g
retarding force= m1*g
F=ma
m2*g-m2*a = tension
at the pulley, one has m1+ m2 handing from it, force=(M1+M2)*g
two masses m1=400.g and m2 =600.g are connected with a light string which goes over a frictionless pulley of negligible mass. the two systems are released from rest. a) what is the acceleration of each mass? b) what is the tension force in the string? and c) what is the support force in the pivot of the pulley?
2 answers
for part c, the support force is double the tension. The tension force is what "pushes" down on the pulley. So, it will be 2T.