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Considering a rack of blocks, and that I am able to hold it up in a horizontal line without it collapsing.

I am trying to draw the free body diagram of one of the blocks in the middle portion, block A.

I would draw two Normal contact forces from the sides, the weight, and two static frictions from the adjacent blocks. Naturally this explains why A is in equilibrium.

However, I get confused when I draw the free body diagram for the adjacent blocks, knowing that there must be an Action Reaction pair of the static friction. (Static friction by B on A = - Static friction by A on B), and this AR of static friction points downwards on B.

Am I getting something wrong here?

Rack of blocks

Urb
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esdoublelef
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5 Answers5

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Yes there is an upward force on $A$ due to $B$ and an equal and opposite reaction force on $B$ due to $A$ but then there is a larger upward force on $B$ due to $C$ and the corresponding reaction force on $C$ and then an even larger upward force on $C$ due to $D$ etc . . . with the largest upward forces being applied externally at the outer ends of the line of blocks.

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As you move away from the centre the upward frictional force must increase because that force is "holding up" all those blocks which are closer to the centre. In the diagram the Newton third law pairs of forces are bracketed.
Each block has a net upward force equal to its weight.

Farcher
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    " then there is a larger upward force on B due to C " But shouldnt the upward force on B due to C and due to A be the same ? Since, we know that B is at rest, so lateral force on B due to A should be same as lateral force on B due to C. And if lateral forces are the same, the friction forces should be the same as well. – silverrahul Apr 14 '21 at 09:36
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    As you move away from the centre the upward frictional force must increase because that force is "holding up" all those blocks which are closer to the centre. Each block has a net upward force equal to its weight. – Farcher Apr 14 '21 at 10:52
  • " Each block has a net upward force equal to its weight. " I disagree. I think it should be that each block experiences upward force equal to its weight + the frictional force exerted downwards by the adjoining block which is closer to the center than it. Look at the diagram in the quesiton. Block B has an additional force downwards apart from its weight. Hence, the upwards force should also increase to account for this extra force – silverrahul Apr 14 '21 at 15:16
  • @silverrahul I have added a diagram to my answer to show that the net upward force due to friction is equal to the weight of a block. – Farcher Apr 14 '21 at 22:55
  • Okay, But you can see that the blocks towards the centre are experiencing less friction force (W/2) than the blocks towards the end (5W/2) . Which means, the end blocks are closer to the friction limit than the center block. Which means if you were to put some weight on top of the boxes, the ones near the ends would be able to support less weight than the ones near the center. But we know from experience, that the central boxes are able to support less weight before giving up than the others. I cant help thinking we are missing something else, and there is something more going on. – silverrahul Apr 15 '21 at 05:33
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Then there is a larger upward force on B due to C " But shouldnt the upward force on B due to C and due to A be the same ? Since, we know that B is at rest, so lateral force on B due to A should be same as lateral force on B due to C. And if lateral forces are the same, the friction forces should be the same as well.

The answer to this question is that when normal reactions are same maximum limiting value of friction will be same but it may happen that for inner blocks friction is less than this maximimum limiting force while for outer blocks it is equal to the maximum limiting force.

Möbius
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Static friction can exert any force on a body within the maximum given by the coefficient of friction, and the actual force is just the amount needed to keep the body motionless.

Think of a body on a (non-slippery) surface slightly inclined to the right, e.g. by 5 degrees. It surely gets less than the maximum static friction force, otherwise friction to the left would be greater than the gravitational component to the right, and the block would start moving to the left, which doesn't happen.

It's correct that, in your setting, the lateral forces onto a block from the left and right are equal, but that doesn't imply identical left and right friction forces, only identical friction force limits.

Ralf Kleberhoff
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  • " the lateral forces onto a block from the left and right are equal, but that doesn't imply identical left and right friction forces, only identical friction force limits. " This is true. But assume that the block A is the exact central one. So, in this case due to symmetry, the left and right frictional forces on it from it's left and right neighbour should be the same, right ? – silverrahul Apr 14 '21 at 10:37
  • @silverrahul Yes, because of symmetry. – Ralf Kleberhoff Apr 14 '21 at 12:17
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Another possibility which I thought about drawing FBD in this question is that there may be a frictional force on A by the block left to it in an upward direction, and fictional force on A by B in a downward direction, and due to action-reaction there is an upward force on B by A. And this friction force along with the weight is balanced by the friction force on B by C and it continues till the end. And this or what you have made is correct depends on the value of the coefficient of friction.

But if you ask my personal views this what I have told you sounds good because it has a fixed pattern that on each block there is an upward as well as downward force but what you have made in that only on A there is both upward frictional force acting on it while except A all has one upward and one downward force acting on it. So the question arises, "Why on the Earth, Nature favors A."

hope this helps you. thank you

lee
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Thank you Farcher for that little hint!

I think my answer is: when we look at A+B+C, the static frictions fs1+fs2 must balance all 3 weights!

But when we look at object B itself, fs3 and fs4 is smaller! (in this case fs3+fs4 is 1/3 that of fs1+fs2)

Then when we look at Object C, there's a fs4' pointing down and there's the fs2 pointing up. fs2 is large enough to balance the downwards fs4'+Wc.

But this also means that the N forces on both sides of C will be different, and not aligned, because we still have to consider balancing rotational moments (caused by the anticlockwise moments of fs4' and fs2).

I hope this is right now!

Rough solution

esdoublelef
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  • The problem with this is that according to this logic, the blocks towards the centre are experiencing less friction force than the blocks towards the 2 ends. Which means, the blocks at the 2 ends are closer to the friction limit than the one at the centre. Which means if you were to put some weight on top of the boxes, the ones near the ends would be able to support less weight ( since they are closer to the friction limit ) than the ones near the center. But i think we can all agree from experience, that the central boxes are able to support less weight before giving up than the others – silverrahul Apr 14 '21 at 15:13