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 0.4×50N=20N. The vertical component of the force at the bottom right corner is Fsin40∘, so the moment of this force is 1.2×Fsin40∘=20N, giving me an F value of 25.9N. This isn't correct, but I don't see where I've gone wrong.
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.
The vertical component of the force at the bottom right corner is Fsin40∘, so the moment of this force is 1.2×Fsin40∘=20N, giving me an F value of 25.9N. 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.
So I have to use both the vertical and the horizontal components of the force? So F(0.8sin40+1.2cos40)=20N?
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).
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?
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.