Thread: tension, acceleration of a weight block

the weight of the block on the table is 468 N and that of the hanging block is 170 N. Ignore all frictional effects, and assuming the pulley to be massless.

Find the acceleration of the two blocks.
m/s^2

Find the tension in the cord.
N


i know the formula Fnet = ma

however they say the weight is 468 and 170, i know these could be added together but, if there the weight then whats the Fnet, or also if they were switched around.

need help please.

also the tension of the cord is the same throughout the system, i also know that.

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Originally Posted by rcmango

the weight of the block on the table is 468 N and that of the hanging block is 170 N. Ignore all frictional effects, and assuming the pulley to be massless.

Find the acceleration of the two blocks.
m/s^2

Find the tension in the cord.
N


i know the formula Fnet = ma

however they say the weight is 468 and 170, i know these could be added together but, if there the weight then whats the Fnet, or also if they were switched around.

need help please.

also the tension of the cord is the same throughout the system, i also know that.

=========================================

You have not one, but two Free-Body Diagrams to deal with. Make one for each block.

Three things to note:
1) The tension on the block that is hanging is equal to the tension on the block on the horizontal surface.

2) When you choose your coordinate systems for each Free-Body diagram, always choose one of the positive directions in the direction of the acceleration of the blocks. This will save you enormous troubles with negative signs. (You do realize the acceleration of both blocks has the same magnitude, right?)

3) When you take the sum of forces you are doing it only for the FBD under consideration. So only use the forces in that FBD. Also the mass is the mass of the object in the FBD, not the total mass.

You will wind up with a system of equations to solve.

-Dan