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Was doing this question on moments, and as part of my working out I labelled a reaction force for the child (question will be attached below). Yet the answers don't mention anything about the child having a reaction force. Why is there no reaction force for the child?

3pointonefour

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Here's the question

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RogerOxon

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(Original post by 3pointonefour)
Was doing this question on moments, and as part of my working out I labelled a reaction force for the child (question will be attached below). Yet the answers don't mention anything about the child having a reaction force. Why is there no reaction force for the child?

The child as a weight, acting directly downwards. The only possible reaction force from the ladder is therefore vertically upwards. However, we're not drawing an exploded diagram, so we only care about the force on the ladder, which is their weight.

The wall is smooth. What can you say about the reaction force there?

What forces act on the bottom of the ladder? What can you say about the reaction force there? (Hint: Consider vertical equilibrium)

3pointonefour

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(Original post by RogerOxon)
The child as a weight, acting directly downwards. The only possible reaction force from the ladder is therefore vertically upwards. However, we're not drawing an exploded diagram, so we only care about the force on the ladder, which is their weight.

The wall is smooth. What can you say about the reaction force there?

What forces act on the bottom of the ladder? What can you say about the reaction force there? (Hint: Consider vertical equilibrium)

I thought that a normal reaction is always perpendicular to the surface?

What do you mean by exploded diagram?

I've done the question, but im asking why - in terms of forces and physics - is there not a reaction force perpendicular (or at all) to the ladder.

Sorry for asking so many questions I'm self-teaching and I find that mechanics in particular leaves lots of gaps so I want to increase my understanding

3pointonefour

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Bump

3pointonefour

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Bump the second

BobbJo

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(Original post by 3pointonefour)
Bump

The child exerts a force on the ladder. This force is called his weight. By Newton's 3rd Law, the ladder exerts a force on the child equal and opposite to the weight. As noted earlier, the force is vertically upwards. This is opposite to the weight.

Care to be taken here: the LADDER exerts a force on the child.

When taking moments, you consider equilibrium for the ladder and only forces on the ladder

Edit: You bumped twice in 4 minutes. Be more patient

3pointonefour

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(Original post by BobbJo)
The child exerts a force on the ladder. This force is called his weight. By Newton's 3rd Law, the ladder exerts a force on the child equal and opposite to the weight. As noted earlier, the force is vertically upwards. This is opposite to the weight.

Care to be taken here: the LADDER exerts a force on the child.

When taking moments, you consider equilibrium for the ladder and only forces on the ladder

Edit: You bumped twice in 4 minutes. Be more patient

Quite confused, if every action has an equal and opposite reaction, how is it that there are resultant forces at all? Shouldn't everything be cancelled out?

So if the reaction force is equal and opposite to the weight, shouldnt they cancel out and I should thus leave it out of my calculations, no?

Sorry about the impatience, I get really annoyed when I just don't understand something in maths.

BobbJo

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(Original post by 3pointonefour)
Quite confused, if every action has an equal and opposite reaction, how is it that there are resultant forces at all?

You need to read the statement more carefully.

Here it is again: The child exerts a force [weight] on the ladder. The ladder exerts an equal and opposite force on the child.

Thus: The forces act on 2 different bodies.

Shouldn't everything be cancelled out?

This is a common misconception. The answer is no as explained above.

So if the reaction force is equal and opposite to the weight, shouldnt they cancel out and I should thus leave it out of my calculations, no?

No they don't cancel out as they don't act on the same bodies.

3pointonefour

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(Original post by BobbJo)
You need to read the statement more carefully.

Here it is again: The child exerts a force [weight] on the ladder. The ladder exerts an equal and opposite force on the child.

Thus: The forces act on 2 different bodies.

This is a common misconception. The answer is no as explained above.

No they don't cancel out as they don't act on the same bodies.

I see, so why do I still not add on a reaction force (of any direction) to the child but when i do questions about particles on an inclined plane, there is a reaction force and it's perpendicular to the plane?

BobbJo

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(Original post by 3pointonefour)
I see, so why do I still not add on a reaction force (of any direction) to the child

I don't understand what you mean by this but I will try to interpret it. In the diagram you include only the forces on the ladder. The reaction is a force the ladder exerts on the child. So it's not included in the diagram

but when i do questions about particles on an inclined plane, there is a reaction force and it's perpendicular to the plane?

Because the normal component of weight is balanced by the reaction force. The normal reaction acts on the body. [There is no motion perpendicular to the plane, so forces must be balanced in that direction]

Last edited by BobbJo; 2 months ago

3pointonefour

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(Original post by BobbJo)
I don't understand what you mean by this but I will try to interpret it. In the diagram you are only the forces on the ladder. The reaction is a force the ladder exerts on the child. So it's not included in the diagram

Because the normal component of weight is balanced by the reaction force. The normal reaction acts on the body. [There is no motion perpendicular to the plane, so forces must be balanced in that direction]

Ok thanks very much for clearing up my misconceptions. You've been a great help!

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