How do I find pulley tension without knowing acceleration?

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smruthi92
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hey guys, please see the attachment for the question.

i drew FBDs and everything. i want to know a few things tho. how do i calculate the tension in the pulley?

if they'd given me acceleration i could have used T-mg =ma. but they havent. also once the Tension is found, do i just find the moment of inertia around the centre of mass and then resolve forces?

thanks,
s.s
 

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hi smruthi92! :smile:

call the acceleration a, and the angular acceleration of the rod α, then do F = ma and τ = Iα :wink:
 
tiny-tim said:
hi smruthi92! :smile:

call the acceleration a, and the angular acceleration of the rod α, then do F = ma and τ = Iα :wink:

hey tim,
i can't seem to find a to find the tension, that's my problem. :(
 
call it a, and you can eliminate it later …

what do you get? :smile:​
 
tiny-tim said:
call it a, and you can eliminate it later …

what do you get? :smile:​

im sorry i don't get it :(
using M = I alpha

I = mL^2/12

but that's all i understand. u know the tension, it produces a moment right? so using that i got:

M = I alpha
T = I alpha
ma +mg = I x alpha
 
smruthi92 said:
ma +mg = I x alpha

where does this come from? :confused:

you need separate equations (both involving T) for the mass and for the rod, and you also need an equation relating a to α

(oh, and your moment of inertia is about the wrong point)

try again! :smile:​
 
tiny-tim said:
where does this come from? :confused:

you need separate equations (both involving T) for the mass and for the rod, and you also need an equation relating a to α

(oh, and your moment of inertia is about the wrong point)

try again! :smile:​

sorry that's right.

I = mL^2/12 + md^2

a = alpha x r

M = I x alpha

M = T d
T = mg +ma.

but u know the a we find from a = alpha x r, is it the same a of the mass on the pulley?
 
hi smruthi92! :smile:

yes, the string stays the same length, so a at one end of the string must be the same (well, minus) as a at the other end of the string :wink:

(and don't forget you must include the torque of the weight of the rod)
 
tiny-tim said:
hi smruthi92! :smile:

yes, the string stays the same length, so a at one end of the string must be the same (well, minus) as a at the other end of the string :wink:

(and don't forget you must include the torque of the weight of the rod)

awesome! thank u so much, just one last question,. after finding the total torque at point A, how do u find the reaction force?
 
take components in the x and y directions and/or moments about the other end :wink:
 
tiny-tim said:
take components in the x and y directions and/or moments about the other end :wink:

ok thank u so much tim! uve helped me so much! :D