Work out force transmitted to sensor

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The discussion centers on calculating the force transmitted to a sensor in a wind tunnel setup, where a drag force of 1.20 N and a lift force of 1.75 N act on a test object. The force transmitted to the sensor at point B is determined to be 2.88 N, calculated using torque principles from the lever system. The participants clarify the importance of considering both horizontal and vertical components of the forces at the pivot point O, leading to equations that sum the forces to zero. The conversation emphasizes the need for accurate free body diagrams and the correct application of torque calculations. Overall, the thread provides insights into resolving forces in mechanical systems effectively.
  • #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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