## magnitude of the torque about the rotation axis at an instant

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

A wheel of diameter 26.0 cm is constrained to rotate in the xy plane, about the z axis, which passes through its center. A force F = ( - 31.0 i + 49.0 j )N acts at a point on the edge of the wheel that lies exactly on the x axis at a particular instant. What is the magnitude of the torque about the rotation axis at this instant?

2. Relevant equations

Torque= RF = RFsin(theta)

Torque= (I)(alpha)

3. The attempt at a solution

I can imagine that I need to integrate the function given for the force or break the force down to its components but I am completely confused with this one.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
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 Recognitions: Gold Member Homework Help Science Advisor Welcome to the Physics Forums! Your 2nd relevant equation is not necessary, since the problem is not asking about angular acceleration. It is just asking for the torque magnitude about the center of the wheel. Your first relevant equation is correct, but you need first to identify R, F, and theta. This is a little involved unless you keep the force in its vector component form, and find the torque produced by each componet, then add them up algebraically. Or it may be simpler for you to look at the torques produced by each x and y component of the force by using the alternate definition of torque: Torque = force times perpendicular distance from the line of action of the force to the pivot point. In general, watch plus and minus signs...if clockwise is plus, counterclockwise is minus.
 So maybe this works? T=sqrt[(RxFx)+(RyFy)] T=sqrt[(.13 x 31.0)^2 + (0 x 49)^2] T =sqrt[16.24] = 4.03

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## magnitude of the torque about the rotation axis at an instant

No.

First, you don't take the sq rt of the sum of the squares for forces in a 2D plane. The torques (T) add up algebraically, since they are about the same axis. For example, If T due to Fx was plus 3, and T due to Fy was minus 4, then T would be -1, not 5.

Second, you are not calculating the distances properly for each component. The point of application of the force lies on the x axis.
 So: Tx = (.13)(-31.0) = -4.03 Ty = (.13)(49.0) = 6.37 So T= Tx+Ty = 2.34? Thanks for helping me through this, by the way. I know this is probably very simple but I really haven't seen any examples like this.

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 Quote by pcml100 So: Tx = (.13)(-31.0) = -4.03
No.
 Ty = (.13)(49.0) = 6.37
Yes!
 So T= Tx+Ty = 2.34?
Correct for the value of Tx, where Tx is the torque due to the x component of the force.
 Thanks for helping me through this, by the way. I know this is probably very simple but I really haven't seen any examples like this.
You have the magnitude of 'Ty' calculated correctly, where Ty is the torque due to the y component of the force.

Now correct your Tx value. The torque due to the x component of the force is the x component of the force times the perpendicular distance from its line of action to the center of the wheel. What is the perpendicualr distance from the line of action of the x component of the force to the pivot point? (you sort of had the right idea earlier but you mixed up your components). Also, it is incorrect to label the torques as Tx and Ty. They are both actually Tz, because the direction of the torques of forces in the x and y direction in an xy plane is along the z axis for both .
 If the point where the force is being applied is at the x axis at that particular instant, I guess that the perpendicular distance on the y axis is 0? In that case, T = 6.37 because there would be no x component (it would =0)?

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