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Hard hard physics, reps

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

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Reply 1
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Lola1244
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oh wait is it just 60 degrees below the horizontal for the direction?
Original post by Lola1244
oh wait is it just 60 degrees below the horizontal for the direction?


Yeah, now give me my rep :wink:
Reply 3
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Original post by ScienceFantatic
Yeah, now give me my rep :wink:


once you tell me how to work out the magnitude :wink:
Reply 4
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Lola1244
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i tried to draw a vector triangle and use the cosine rule
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
Reply 6
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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
Original post by Lola1244
do u not have any idea how to go abut it


Sorry I'm still GCSE level
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?
Reply 9
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Original post by CharlieGEM
Is the string at C?


no the wire is at B which is the midpoint of the rod
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?
Reply 11
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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
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.
Reply 13
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Lola1244
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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
Reply 14
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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
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.
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.
Reply 17
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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
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.
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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