# Equilibrium applications of newton's law of motion II

Box 1 is resting on the table with box 2 resting on top of box 1. a massless rope passes over a massless, frictionless pulley. One end of the rope is connected to box and the other end is connected to box 3. The weights of the three boxes are w1=55.0N, W2=25.7N. Determine the mangitude of the normal force that the table exerts on box 1.

So i drew it out and the tension and normal force would both go up and the weight would go the opposite direction...
So would I add (M*g)+tension? then subtract the weight? where would box three come into play??
Thank you

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I don't think the tension has anything to do with this... Unless I'm misunderstanding this problem, box 1 is flat on a table with 2 on top of 1, and box 3 is suspended in the air by a rope-pulley system. So the weight of box 3 would create the tension in the rope for the (box1/box2) weight. However, in this problem, the tension is parallel to the table, whereas we are looking for the normal force which is always perpendicular to the surface. So my best guess would be that the normal would just be (mass of box1+mass box2)g since the box is being held still on the table. If i'm misunderstanding the problem, let me know.

So i did what you said and (55+38.8)9.8 and got 919.24 but it was wrong so I'm assuming that box 3 would create upward tension?? since it is hanging??

well the thing is that the weight of the box is a downward force and it is canceled out by the tension in the rope which is an upward force, so that should be in equilibrium assuming that the box is not in motion

maybe i'm still misunderstanding, is there a way you could attach a picture of the system?

Ahhh i think I see where your problem is, you added the WEIGHTS together and multiplied by g, you needed to add the MASSES together and multiply by g. Luckily this problem is easier than you thought, all you need to do is add the two weights together since weight=mass*g. That should give you your normal force.