Why Did My Torque Sign Change in Rotational Equilibrium?

[JoeyBob]

Homework Statement

See attached

Relevant Equations

0=r x F

See attached for work. I did notice that making the torque from the force of gravity negative I got the right answer, but don't understand what I did wrong (its positive in my solution). i hat cross negative k hat is a positive number after all.

 

[Hamiltonian]

I find it rather hard to understand your attempt at a solution could you use LaTeX or type the solution?
(if not maybe use a scanning app to scan your work and then upload it)
https://www.physicsforums.com/help/latexhelp/

 

[JoeyBob]

I find it rather hard to understand your attempt at a solution could you use LaTeX or type the solution?
(if not maybe use a scanning app to scan your work and then upload it)
https://www.physicsforums.com/help/latexhelp/

Sure I'll type it...

T=Friction force = umg = 0.47*6.6*9.8 = 30.3996 N (magnitude)

0=Torque from Mass + Torque from rope

0= (2/3*1.8 *cos(angle) i hat) x (-mg k hat) + (1.8*sin(angle) k hat) x (-T i hat)

0=1.2mgcos(angle)-1.8Tsin(angle)

0=77.616cos(angle)-54.71928*sin(angle)

 

[Delta2]

Without having to resolve this using the cross product of the unit vectors, can you see that the weight torque tends to rotate the bar clockwise, while the tension from rope torque tends to rotate the bar counter clockwise. So you can't take both as positive or both as negative, if you take one torque as positive the other torque must be negative and vice versa.

 

[JoeyBob]

Without having to resolve this using the cross product of the unit vectors, can you see that the weight torque tends to rotate the bar clockwise, while the tension from rope torque tends to rotate the bar counter clockwise. So you can't take both as positive or both as negative, if you take one torque as positive the other torque must be negative and vice versa.

I got it thanks. Figuring out angles confuses me often. Drawing a diagram does help me.

 

[JoeyBob]

Without having to resolve this using the cross product of the unit vectors, can you see that the weight torque tends to rotate the bar clockwise, while the tension from rope torque tends to rotate the bar counter clockwise. So you can't take both as positive or both as negative, if you take one torque as positive the other torque must be negative and vice versa.

If you don't mind me asking, why isn't friction force part of the rotational equilibrium? Or the normal force? Is it because theyre at the origin?

 

[Delta2]

If you don't mind me asking, why isn't friction force part of the rotational equilibrium? Or the normal force? Is it because theyre at the origin?

Because you chose to take moments around the origin (the point at the ground) and therefore the moment of friction and the moment of the normal force is zero because the distance of the point of application of these forces, to the origin is zero (in rxF r is zero for this forces). If you choose to take moments around another point then you would have to include the moments of the normal force and of the friction force because they wouldn't be zero

 

[JoeyBob]

Because you chose to take moments around the origin (the point at the ground) and therefore the moment of friction and the moment of the normal force is zero because the distance of the point of application of these forces, to the origin is zero (in rxF r is zero for this forces). If you choose to take moments around another point then you would have to include the moments of the normal force and of the friction force because they wouldn't be zero

Okay I am still a bit confused with the tension... I understand how one is negative and one is positive because theyll rotate the beam in opposite directions, but cross product wise this doesn't always work out for me.

Take the attached question for instance. In this case Tension is in the positive i hat and force of gravity in the negative k hat. The angle gives (cos(angle) k hat + sin(angle) i hat). When multiplying by the forces both will be positive, which doesn't make sense because theyre pulling in opposite directions.

 

[Delta2]

Actually the arm is $$\vec{r}=\cos\theta \hat k-\sin\theta \hat i$$
so the moment of T is $$\vec{r}\times \vec{T}=T\cos\theta\hat k\times \hat i+0$$ while the moment of weight is $$\frac{1}{2} \vec{r} \times \vec{B}=-B\sin\theta \hat i \times -\hat k=B\sin\theta\hat i\times \hat k$$

So they have opposite signs because of the anticommutative property of the cross product it is $$\hat i\times\hat k=-\hat k\times\hat i$$.

 

[JoeyBob]

Actually the arm is $$\vec{r}=\cos\theta \hat k-\sin\theta \hat i$$
so the moment of T is $$\vec{r}\times \vec{T}=T\cos\theta\hat k\times \hat i+0$$ while the moment of weight is $$\frac{1}{2} \vec{r} \times \vec{B}=-B\sin\theta \hat i \times -\hat k=B\sin\theta\hat i\times \hat k$$

So they have opposite signs because of the anticommutative property of the cross product it is $$\hat i\times\hat k=-\hat k\times\hat i$$.

Oh, so its -sin(angle) because it goes to the left. Okay I think I understand it all now. Thanks for all the help.