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My solution


Let me explain

$n_{bg}$ means the force exerted by the block on the ground. This is because the normal force of the slab exerts (or touches) a force on the block and by Newton's Third Law, it repels with a force n. So $n_{ground}$ is actually a force exerted by the ground and the force exerted by the ground on the block

Also I kinda intuitively knew to put a friction force on the slab on the right. Is this a consequence of Newton's third Law? As the block slides to the right, the friction force drags it from the left. On the slab, (imagining now) wouldn't the heat/friction be sliding on top from the left?

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in reality your N(bg) is N(action-reaction force) –  sum2000 Dec 19 '11 at 20:21
I know, but the original solutions don't have that. –  Hawk Dec 19 '11 at 20:27
it depends on how you take, some books follow different conventions, although the normal way of doing it by action-reaction, both are same friend. –  sum2000 Dec 19 '11 at 20:33
I don't follow. The original solution is MISSING one force. –  Hawk Dec 19 '11 at 20:35
As explained in orion's answer, the original solution is correct. A couple of the labels are misleading IMO but it does show the correct set of forces. If you're talking about $n_{bg}$ being missing, that's because the block does not actually exert a force on the ground. Only objects in contact can exert forces on each other, except or gravity. –  David Z Dec 20 '11 at 8:13

2 Answers 2

up vote 4 down vote accepted

The Free body diagram (FBD) given in solution is correct. Your FBD has one error.

The arrows on the dot (which is a simplistic representation of the objects) in a FBD is a representation of force acting on that object. Here, all arrows that you have drawn on the dot that represents slab, are the forces acting on the slab. So, you cannot draw $n_{bg}$, which as you named is force exerted by block on ground, as an arrow on slab.*

Action and reaction force are always acting on two separate bodies, it is very important. Here are the action reaction pairs. I am using your nomenclatures.

(a) Action Force: Block applies a force $F_{bg}$ on slab.

Reaction force: Slab applies a normal force $n$ on block.

(b) Action force: Due to weight of block and slab, slab applies a force on ground. $F_{bg} + F_{sg}$

Reaction force: Ground applies a normal force on slab. $n_{ground}$

(c) Action force: Block applies a force on slab due to friction towards right. $f$

Reaction force: Slab applies a force on block due to friction towards left. $f$

I hope it will be clear, now.

  • As David has rightly pointed out, block does not apply force on ground, rather it applies a force on slab, as it is in contact with it. I am ignoring gravitational pull of block on ground (or Earth).
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I am a little confused. Suppose I have two blocks, doesn't matter what their mass is. They are hanged up by a vertical distance of x meters. Assume no air resistance for now. Then the first block is dragged by gravity. The second block is also dragged by gravity. But since on top of the second block, the gravitational force from the first block comes down, does that gravitational force exert on the second block? –  Hawk Dec 19 '11 at 23:49
@ Jak: I am not able to interpret what exactly you are asking, hence responding as I have understood the meaning of your comment. When you draw arrows in FBD, arrow head is just showing you the direction, the tail is important as it is showing you on what point force is acting. Hence, gravitation pull on upper block will not act on the top of the second block, it points downwards to show direction, it does not say that as it is acting on the second block. Kindly let me know, if I need to be more clear. I can list down the details of forces acting on both the blocks. –  orion Dec 20 '11 at 13:35

AS action reaction pairs cancel out within one chain, it is custom to simplify the description to only those forces at the beginning and end of the chain.

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