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1. The diagram shows a uniform rectangular sign weighing 50N which is attached to hinge O. The lengths of the rectangle are 0.8m across the top and 1.2m top to bottom. The sign is held steady by the application of a force to the bottom right corner at 40 degrees to the horizontal. Find the size of this force F.

Now the moment of the weight is . The vertical component of the force at the bottom right corner is , so the moment of this force is , giving me an F value of . This isn't correct, but I don't see where I've gone wrong.
2. you can imagine the force being applied to a rod connected directly to the pivot, which should make it easier for you to work out compared rather than thinking about the the rectangle.
3. (Original post by Julii92)
The vertical component of the force at the bottom right corner is , so the moment of this force is , giving me an F value of . This isn't correct, but I don't see where I've gone wrong.
You've worked out the vertical component of the force correctly, however that is not it's moment about O (why did you multiply by 1.2?).

Also, what about the horizontal component of the force, that also has a moment about O.
4. (Original post by ghostwalker)
You've worked out the vertical component of the force correctly, however that is not it's moment about O (why did you multiply by 1.2?).

Also, what about the horizontal component of the force, that also has a moment about O.
So I have to use both the vertical and the horizontal components of the force? So ?
5. (Original post by Julii92)
So I have to use both the vertical and the horizontal components of the force? So ?
Yes, and the S.I. units are Nm, not just N (assuming you're refering to the units of the equation, and not just the force, as that doesn't make sense there).
6. (Original post by ghostwalker)
Yes, and the S.I. units are Nm, not just N (assuming you're refering to the units of the equation, and not just the force, as that doesn't make sense there).
Right, thank you.

Is it possible to use the more direct method of working out the shortest distance between the line of force and the hinge?
7. (Original post by Julii92)
Right, thank you.

Is it possible to use the more direct method of working out the shortest distance between the line of force and the hinge?
I can't see it off the top of my head, but if you can see how to get that distance, then yes.
8. (Original post by ghostwalker)
I can't see it off the top of my head, but if you can see how to get that distance, then yes.
That's my problem. I can't see at all how to get the distance.
9. (Original post by Julii92)
That's my problem. I can't see at all how to get the distance.
Well, I can see a method, but it's a pig to draw and label the diagram properly.

Hope this will help - see attached.

You're looking for the length of the red line, it's made up of 2 parts, a and b.

If you mark in angles you know then hopfully you can see
a is 0.8 cos 50, which is 0.8 sin 40 and
b (the same lenght as b too) is 1.2 cos 40
Attached Images

10. (Original post by ghostwalker)
Well, I can see a method, but it's a pig to draw and label the diagram properly.

Hope this will help - see attached.

You're looking for the length of the red line, it's made up of 2 parts, a and b.

If you mark in angles you know then hopfully you can see
a is 0.8 cos 50, which is 0.8 sin 40 and
b (the same lenght as b too) is 1.2 cos 40
Thanks. What did you use to draw that? MS paint?
11. (Original post by Julii92)
Thanks. What did you use to draw that? MS paint?
Yep!
12. (Original post by ghostwalker)
Yep!
Whenever I draw on MS paint ends up looking like garbage you must be better at it!
13. I was wondering,what is the name of the horizontal(left) force that makes the object require a right horizontal force to maintain equilibrium?

This was posted from The Student Room's iPhone/iPad A
I was wondering,what is the name of the horizontal(left) force that makes the object require a right horizontal force to maintain equilibrium?
It has no specific name that I am aware of, other than being the horizontal component of the reaction at the pivot, O.
15. Huh isnt reaction force upwards (towards me if its 3D) instead of left?

This was posted from The Student Room's iPhone/iPad A
Huh isnt reaction force upwards (towards me if its 3D) instead of left?
If it's to hang in the position given, the reaction force at O, will have an upwards component and a component to the left.

17. Reaction force shouldnt it be only perpendicular to the surface of contact hence named normal reaction force? Then the horizontal leftward force idk..think it's due to the weight but weight is straight vertical downwards.man,what is that force called (((

This was posted from The Student Room's iPhone/iPad A
Reaction force shouldnt it be only perpendicular to the surface of contact hence named normal reaction force? Then the horizontal leftward force idk..think it's due to the weight but weight is straight vertical downwards.man,what is that force called (((
The only contacts here are a hinge at O, and the applied force at the bottom right corner; there is no surface, per se.

A reaction force is only normal to a surface, if the applied force is normal to that surface. If you had a frictional component, then the reaction force, won't be normal, e.g A ladder against a smooth wall, on rough ground. The reaction force at the base of the ladder is not normal to the ground.

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