# AQA M2B moments question!

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#1
Hi, I was wondering if there's anyone who could explain to me how reaction forces work.

Sometimes it acts perpendicular to the wall, sometimes it acts perpendicular to the rod and I'm a bit confused as to understanding why.

One example is from the past paper and if you could help me using question 9, that would be extremely helpful.

Thank you!
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3 years ago
#2
The best way to think about it is that the normal reaction force acts perpendicularly to the point of contact between the object you are interested in and what it is in contact with.

In the case of ladders, they are in contact with an entire horizontal or vertical serface, so the reaction force will be perpendicular to that surface, as that is where the contact is. In the case of Q9 on your attached paper, the contact is at a single point (if we are only looking at a 2D cross-section), so the reaction force will be perpendicular to the contact between the rod and that point - so perpendicular to the rod.

The thing that I tend to find confusing (and this happens more in physics than maths) is where the reaction force is described as being the total force that an object feels because of it's contact with the ground (for example), so the resultant of what we are (correctly!) calling the normal reaction force and any frictional contact. It seems so much clearer and easier to deal with these parts separately.
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