Recent content by ThunderboltGT

Okay, I see that I was supposed to add (m)(-a)= \Sigma F_y Thanks for the help.

\Sigma F_x = 0 + T = (m)(a) so T= (m)(a). I plug that into \Sigma F_y= T - W_2= (m)(-a) and come out with \Sigma F_y= (m)(a) - W_2= (m)(-a) and substitute W for (m)(g). Now the equation looks like (m)(a) -(m)(g). (m)(a)= (m)(g) then divide by m and a=g, that's what I keep getting.

Okay, for \Sigma F_x= -Wsin(180) + T. I think that is right because F_N is a force that the table exerts up in the +y direction on Block A while W is points in the -y direction. The tension is moving Block A in the +x direction. Acceleration is the same for both blocks because the rope...

Homework Statement Two blocks, each of mass m, are connected by a string of constant length 4h. Block A is placed on top of a table 3h from the side and Block B hangs over the edge. The tabletop is 2h above the floor. At time t=0, Block B is released from rest at a distance 1h. Use variables...

Thanks! I'm going to try it out.