two blocks on a pulley, find tension and acceleration

Oct 2009
In the drawing, the weight of the block on the table is 422 N and that of the hanging block is 185 N. Ignoring all frictional effects and assuming the pulley to be massless, find (a) the acceleration of the two blocks and (b) the tension in the cord.

(in the diagram there is a block on a table attached to a string, with a pulley at the corner of the table and on the other side of the string the 185N block hanging away from all surfaces)

My work:
Tension is the same for both blocks, much like acceleration is the same magnitude but opposite sign (a2 = -a1)

sum of forces on block 1 in the x direction (since Fn and Fg cancel out)
= T = m1a1

sum of forces on block 2 in the y direction (only Tension and Fg acting)
= T - Fg = m2a2y

since Tensions are the same, we can equate these two equations to:
m1a1 - Fg = m2a2y

This is where i get stuck!!
I need to solve for a then solve for T
but how can i solve for a2 (which is the -a1) when they will cancel?
I'm really lost on the algebra on this one, please help

all thanks


PHF Helper
Feb 2009
As i have mentioned in earlier posts also, i am not comfortable with the convention of assigning signs to directions because i lose the "physical feel" of the problem.

a1 and a2 have the same magnitude but different directions . They are at right angles actually so a1 is not = -a2

Work with the magnitude of the accelerations. Treat both as having the value a and then solve. The directions will then be horizontal for 422 and vertically downwards for 185.

Remember these values of 422 and 185 are their weights. You have to divide them by g to get the value of the masses

Similar problems have been solved in this section . Do look them up for further details