Finding torque

Homework Statement

What is the torque τB about axis B due to the force F⃗ ? (B is the point at Cartesian coordinates (0,b), located a distance b from the origin along the y axis.)
Express the torque about axis B in terms of F, θ, ϕ, π, and/or other given coordinate data.

[/B]
τ = RF sin (θ)

The Attempt at a Solution

So I know that the answer is Tb = bFsin (pi/2+θ)
What I don't understand is how to get pi/2+θ?
I know that θ is the angle between r vector and F vector and since that's the case shouldn't the answer be pi/2-θ?

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Orodruin
Staff Emeritus
Homework Helper
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Can you please show us your own effort? What did you get?

Can you please show us your own effort? What did you get?
Well I thought the answer was Tb = bFsin (pi/2-θ) but that's obviously wrong. I would like to know why. My homework says that alpha is the angle between r vector and F vector therefore I thought that it would be pi/2-θ.

Orodruin
Staff Emeritus
Homework Helper
Gold Member
The angle ##\alpha## is drawn in the figure. Note that the vector ##\vec r## is the vector from the point relative to which you want to know the torque to the point where the force acts. In this case the vector from B to A.

Edit: Also note that, since what matters is the component of ##\vec F## perpendicular to ##\vec r##, it does not matter if you use ##\alpha## or ##\pi - \alpha## since
$$\sin(\pi - \alpha) = \sin(\pi)\cos(\alpha) - \cos(\pi)\sin(\alpha) = 0 + \sin(\alpha) = \sin(\alpha).$$
It holds that
$$\sin(\pi/2 - \theta) = \sin(\pi/2)\cos(\theta) - \sin(\theta)\cos(\pi/2) = \cos(\theta)\sin(\pi/2) = \cos(\theta)\sin(\pi/2) + \sin(\theta)\cos(\pi/2) = \sin(\pi/2 + \theta).$$