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# Hard hard physics, reps

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1. A light rigid straight beam ABC is smoothly hinged at end A on a vertical wall and supported in horizontal position by a wire which is attached to the wall and midpoint of beam b at an angle of 60degrees to the beam.
Calculate the tension in the wire and direction and magnitude of the force on the hinge when a mass M is suspended from the string

okay so I took moments about point A
Tsin60=2Mg
so T is 2Mg/sin60... T is 2.3MG

i dont get how to get the force in the wire
2. oh wait is it just 60 degrees below the horizontal for the direction?
3. (Original post by Lola1244)
oh wait is it just 60 degrees below the horizontal for the direction?
Yeah, now give me my rep
4. (Original post by ScienceFantatic)
Yeah, now give me my rep
once you tell me how to work out the magnitude
5. i tried to draw a vector triangle and use the cosine rule
6. (Original post by Lola1244)
i tried to draw a vector triangle and use the cosine rule
You might get more responses if you post this in the maths forum even though it's physics
7. (Original post by ScienceFantatic)
You might get more responses if you post this in the maths forum even though it's physics
do u not have any idea how to go abut it
8. (Original post by Lola1244)
do u not have any idea how to go abut it
Sorry I'm still GCSE level
9. (Original post by Lola1244)
A light rigid straight beam ABC is smoothly hinged at end A on a vertical wall and supported in horizontal position by a wire which is attached to the wall and midpoint of beam b at an angle of 60degrees to the beam.
Calculate the tension in the wire and direction and magnitude of the force on the hinge when a mass M is suspended from the string

okay so I took moments about point A
Tsin60=2Mg
so T is 2Mg/sin60... T is 2.3MG

i dont get how to get the force in the wire
Is the string at C?
10. (Original post by CharlieGEM)
Is the string at C?
no the wire is at B which is the midpoint of the rod
11. (Original post by Lola1244)
no the wire is at B which is the midpoint of the rod
The string that you hang the mass M from?
12. (Original post by CharlieGEM)
The string that you hang the mass M from?
no no that is at C sorry!
what I did was take moments about A first
13. (Original post by CharlieGEM)
The string that you hang the mass M from?
Because if it is just equate horizontal and vertical forces on the rod.
14. let the length of the rod be 2L
so the perpendicular distance of T from point A is Lsin60
moment of T is therefore TLsin60
this must be equal to moment of M which is 2LMg

so 2LMg=TLsin60
2Mg=Tsin60
15. (Original post by CharlieGEM)
Because if it is just equate horizontal and vertical forces on the rod.
yeah I know but there is a force at point A which is not horizontal or vertical, it must be 60degrees below the horizontal as forces in equilibrium will pass through the same point
16. (Original post by Lola1244)
no no that is at C sorry!
what I did was take moments about A first
So horizontal force in hinge is 2Mg/tan60

And vertical is Mg.
17. (Original post by Lola1244)
yeah I know but there is a force at point A which is not horizontal or vertical, it must be 60degrees below the horizontal as forces in equilibrium will pass through the same point
Both horizontal and vertical total forces must be zero.
18. (Original post by CharlieGEM)
Both horizontal and vertical total forces must be zero.
OH thank you!! haha silly me, how would you get the angle though?
is it 60? please say its 60 hahah
19. (Original post by Lola1244)
OH thank you!! haha silly me, how would you get the angle though?
is it 60? please say its 60 hahah
Nope, arctan(tan60/2) sorry.
20. (Original post by Lola1244)
OH thank you!! haha silly me, how would you get the angle though?
is it 60? please say its 60 hahah
Angle is just arctan(vertical/horizontal).

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