Work out force transmitted to sensor

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SUMMARY

The discussion focuses on calculating the force transmitted to a sensor in a wind tunnel setup involving a lever system. The drag force is 1.20 N, and the lift force is 1.75 N. Using torque calculations, the force transmitted to the sensor at point B is determined to be 2.88 N. Additionally, the horizontal and vertical components of the force at the pivot point O are calculated, with the horizontal component being -2.88 N and the vertical component being -4.63 N.

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  • #31
Thought it might be worth checking that the torque really do sum to zero. If clockwise is +ve then...

Summing the torques about the pivot point gives..

(FD * 1.2) - (FS* 0.5) = ?
(1.2 * 1.2) - (2.88 * 0.5) = ?
1.44 - 1.44 = 0

So that's ok.

However if it's not accelerating in rotation you can actually sum the torques about any point and it will be zero. eg It doesn't have to be a real pivot.

Suppose we pick the sensor and sum the torque about that point. The equation is..

(FD*1.2) + (FL*0.5) +(FY*0.5) = ?

FX and FS don't feature because they pass through the point where the sensor is located (eg tangential distance = 0).

Substitute values..

(1.2*1.2) + (1.75*0.5) + (-4.63*0.5) = ?

1.44 + 0.875 - 2.315 = 0

This ability to pick any point about which to sum torques is handy. For example suppose there is an unknown force somewhere. If you pick the point where that force acts you eliminate it from the equation at a stroke.
 
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  • #32
We learned about this in my lectures this semester and I couldn't visualise it too well, but you've helped me understand it a lot better thank you!
 

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