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I'm trying to figure out the force exerted on blocks positioned on an angled rail due to gravity. This is the scenario I have (apologies for the poor graphic):

enter image description here

where each block (red square) is equipped with a wheel (black circle) which is fitted on a rail (black lines). The top section of the rail is angled at 50 degrees (or just $\theta$) and the bottom section is vertical. I'm trying to figure out on the force exerted on the bottom wheel (green arrow). Here is what I have so far:

We have 3 forces: 1) The force exerted by the top block parallel to the rail is $mg\sin(\theta)$; and 2) The force exerted by the middle and bottom blocks are $mg$. I end up with the following vector triangle:

enter image description here

$R$ is the force exerted on the wheel. Am I missing something?

Any advice appreciated.

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In this situation, the vertical component of the (normal) force of the track acting on the upper block supports part of the weight of the upper block. The problem is to find that normal force. To keep the upper block from rotating there must be a horizontal friction force from the block below acting on the bottom of the upper block. That counters the torque caused by the vertical force from the lower block. To find the normal force I'm going to take the torques about the center of where the blocks meet. I will assume that each block has a width, x, that the upper blocks overlap by a distance, d, that the two upper centers are separated by a distance of, 2L, (center to center), θ = $50^o$, and the wheels are very small. Then the torques about the center point are produced only by the normal and gravity. NL – mg(x/2 – d/2). But (x/2 – d/2) = Lcos(θ). Together these give N = mgcos(θ). (I'm not allowed to give a complete answer.)

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I think the force exerted on the bottom wheel is just the vertical component of your force diagram i.e. $2mg + mg \sin^2 (50^o)$.

The top block exerts a force $mg \sin (50^o)$ on the middle block along the line of the upper part of the track, but the horizontal component of this is opposed by an equal and opposite horizontal force from the lower part of the track. This leaves its vertical component, which is $mg \sin^2 (50^o)$. The middle block can only exert a force on the bottom block along the line of lower part the track, which is vertical, so there are no horizontal forces on the bottom block.

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  • $\begingroup$ Oh, right. But what if the upper part of the tracks were extended like so imgur.com/1p8H8rg Would the force be $R$ in my triangle then? $\endgroup$
    – John M.
    Sep 12 '20 at 8:59
  • $\begingroup$ @JohnM. Now the middle block exerts a force $2mg \sin 50^o$ along the line of the top part of the track on the bottom block, so the force on the bottom wheel is the vector sum of this force and the weight of the bottom block. $\endgroup$
    – gandalf61
    Sep 12 '20 at 9:31

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