## The sum of torques is not the sum of torques?

1. The problem statement, all variables and given/known data

Solutions

3. The attempt at a solution

I don't understand the solutions at all, why do they only take consider one torque at a time? I know the Math works out, but I don't understand why at all

Part b)

Mg - T = Ma

R(Mg - T) = Iα

R(Mg - T) = ˝MR˛(a/R)

R(Ma) = ˝MR˛(a/R)

Ma = ˝Ma

˝Ma = 0

Also, for the second alternative answer, how come it says mg acting on the rim? I thought mg only acts on the center of mass.

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 Blog Entries: 7 Recognitions: Gold Member Homework Help Read and understand the answer. The first method says "Torque about the center of mass" and the second solution says "Torque about the rim". What do these mean?

 Quote by kuruman Read and understand the answer. The first method says "Torque about the center of mass" and the second solution says "Torque about the rim". What do these mean?
Yeah, I don't know what that means, I mean it make more sense if it was the other way around.

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## The sum of torques is not the sum of torques?

What is the definition of a torque? Look in your textbook.
 I mean it make more sense if it was the other way around.
What way would that be?

 Quote by kuruman What is the definition of a torque? Look in your textbook. What way would that be?
I still don't understand, torque = r cross F

Mentor
 Quote by flyingpig I still don't understand, torque = r cross F
That r is the vector from the axis of rotation to the point at which the force is applied. When the axis of rotation is considered to be passing through the center of mass, what is the r vector for the gravitational force? When the axis of rotation is considered to be passing through the contact point on the rim, what is the the r vector for the tensile force?

 No, my problem is the torque they used is only one of the torque acting, it isn't the sum of the torques acting on the body.
 Mentor In both cases they are dealing with the sum of the torques. In both cases, one of those torques is identically zero.

 Quote by D H In both cases they are dealing with the sum of the torques. In both cases, one of those torques is identically zero.
Ohh okay I get it, the mg force is acting on the center of mass, which also happens to be the rotation axis and if we take that as the rotation axis, the tension will obviously have no torque, okay make sense

thanks you two