Help drawing freeform diagram (pulley problem)

AI Thread Summary
A 20 kg block (m1) is sliding on a 10 kg block (m2) on a 20º slope, connected by a massless string over a pulley, with all surfaces being frictionless. The tension in the string acts in the same direction for both blocks, as strings can only pull, not push. The 20 kg block's weight contributes to the normal force on the 10 kg block, increasing the normal force it experiences from the slope. However, since all forces acting on the blocks are perpendicular to the slope, they do not affect the blocks' movement along the incline. Understanding these force interactions is crucial for accurately drawing the freeform diagram.
wraithetc
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"A 20 kg block, m1, is sliding on a 10 kg block, m2. The blocks
are on a 20º slope and are connected by a light string looped
over a pulley. "

All surfaces are frictionless and the string/pulley are massless.

I understand that the 20kg block should slide down since it should have a greater force.

When I draw the diagram I have three forces for each block. The normal force perpendicular to the incline, weight pointing straight down and the tension from the String.

I'm however confused on certain aspects of the diagram.

For instance Tension should always be along the string... so does that mean the tension vector for each block should be in opposite directions?

Also since the 20kg block is sitting a top the 10kg block does that not add to the magnitude of the 10 kg's block weight force?

Thanks for any help in advance.
 
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wraithetc said:
When I draw the diagram I have three forces for each block. The normal force perpendicular to the incline, weight pointing straight down and the tension from the String.
OK. (But the bottom block will experience two normal forces.)

I'm however confused on certain aspects of the diagram.

For instance Tension should always be along the string... so does that mean the tension vector for each block should be in opposite directions?
No. (Strings can't push. :wink:)

Also since the 20kg block is sitting a top the 10kg block does that not add to the magnitude of the 10 kg's block weight force?
It adds to normal force applied to the 10 kg block.

Hint: Consider forces parallel to the incline.
 
wraithetc said:
For instance Tension should always be along the string... so does that mean the tension vector for each block should be in opposite directions?
No. strings can only pull and not push. The strings are obviously pulling the same way. I think you try to use action = reaction here and forget the forces that the pulley exerts on the strings.

Also since the 20kg block is sitting a top the 10kg block does that not add to the magnitude of the 10 kg's block weight force?

since the surfaces of the blocks are frictionless, the force of the big block on the small block will be perpendicular to the slope. It will increase the force that the small block exerts on the slope, and therefore it will increase the normal force that the small block feels, but all these forces are perpendicular to the slope, and the strings and won't affect the movement of the blocks
 
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