# A2 Mechanics - Moments question

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

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?

0

reply

Report

#3

(Original post by

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**)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 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)

0

reply

(Original post by

The child as a weight, acting directly downwards.

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)

**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)

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

0

reply

Report

#7

(Original post by

Bump

**3pointonefour**)Bump

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

0

reply

(Original post by

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

**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

**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.

0

reply

Report

#9

(Original post by

Quite confused, if

**3pointonefour**)Quite confused, if

**every**action has an**equal and opposite**reaction, how is it that there are resultant forces at all?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?

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?

0

reply

(Original post by

You need to read the statement more carefully.

Here it is again: The

Thus: The forces act on

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.

**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.

0

reply

Report

#11

(Original post by

I see, so why do I still not

**3pointonefour**)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?

Last edited by BobbJo; 1 year ago

1

reply

(Original post by

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]

**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]

1

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top