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# Mechanics - What is wrong with my resolving? (forces)

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1. I resolved in two directions and get different S value in each of them. The bottom S value is correct btw, why is the top one wrong?

2. (Original post by Bealzibub)
I resolved in two directions and get different S value in each of them. The bottom S value is correct btw, why is the top one wrong?
You've learnt that resolving in mutually perpendicular directions is really useful, because if you resolve horizontally, then no vertical forces affect anything. If you resolve parallel then no perpendicular forces affect, etc...

In your top calculation, you've resolved perpendicular to the rod, but failed to take into account the friction force. The friction force does not act parallel to the rod and hence you cannot just ignore it.

The bottom calculation, when you resolved vertically, the friction is indeed horizontal and hence mutually perpendicular to the direction you're resolving in, so your calculation works out fine.
3. (Original post by Zacken)
You've learnt that resolving in mutually perpendicular directions is really useful, because if you resolve horizontally, then no vertical forces affect anything. If you resolve parallel then no perpendicular forces affect, etc...

In your top calculation, you've resolved perpendicular to the plane, but failed to take into account the friction force. The friction force does not act parallel to the plane and hence you cannot just ignore it.

The bottom calculation, when you resolved vertically, the friction is indeed horizontal and hence mutually perpendicular to the direction you're resolving in, so your calculation works out fine.
Why is friction not parallel to the plane? Isn't friction always parallel to the plane?

*nvm I thought you meant the plane on which the object was on.
4. (Original post by oShahpo)
Why is friction not parallel to the plane? Isn't friction always parallel to the plane?
This is a rod with an end lying on a rough ground. Left unchecked, the rod will slip horizontally, friction opposes motion and acts in the opposite horizontal direction.
5. (Original post by Zacken)
This is a rod with an end lying on a rough ground. Left unchecked, the rod will slip horizontally, friction opposes motion and acts in the opposite horizontal direction.
Yea I understand. I thought you meant the rod's frictional force was not parallel to the horizontal plane, which is the actual physical plane on which the rod is.
6. (Original post by oShahpo)
Yea I understand. I thought you meant the rod's frictional force was not parallel to the horizontal plane, which is the actual physical plane on which the rod is.
I can see how that would cause confusion, edited my answer to clarify.
7. (Original post by Zacken)
You've learnt that resolving in mutually perpendicular directions is really useful, because if you resolve horizontally, then no vertical forces affect anything. If you resolve parallel then no perpendicular forces affect, etc...

In your top calculation, you've resolved perpendicular to the rod, but failed to take into account the friction force. The friction force does not act parallel to the rod and hence you cannot just ignore it.

The bottom calculation, when you resolved vertically, the friction is indeed horizontal and hence mutually perpendicular to the direction you're resolving in, so your calculation works out fine.
Thanks, well explained
8. (Original post by Bealzibub)
Thanks, well explained
No problem.

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