Torque produced by friction when braking in a car

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When a car brakes, friction generates torque, causing the rear to lift, but the net torque remains zero due to no rotational motion. The discussion highlights that calculations involve forces at specified distances, but questions arise about whether to consider the angle of force application. It is clarified that torque can be calculated using either the perpendicular force component or the moment arm method, both yielding the same result. Understanding these calculations is crucial for analyzing rotational motion in vehicles. The explanation resolves the confusion regarding the role of angles in torque calculations.
jimkers
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Hi

So I am busy studying rotational motion and there's an example in the book where they explain that when a car brakes the force of friction acts as a torque force which explains why a car lifts up at the back when braking.
20180120_171738[1].jpg

Because the car does no rotational motion they say the net torque is zero and that force N1 counteracts the counterclockwise rotation and the rest count towards the counterclockwise rotation.

Here's my problem, when the calculations are being made, they put in the values for all the forces, multiplied by the distance shown on the figure: 1,5 m and 0,75 m. Shouldn't they account for the angle the force makes with the line between the axis of rotation and the tires (point where the force acts)? I thought only the force acting perpendicular to this line serves in this net force equation.

Thanks in advance.
 

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jimkers said:
Shouldn't they account for the angle the force makes with the line between the axis of rotation and the tires (point where the force acts)?
That's true, but there are several equivalent ways of calculating the torque.

jimkers said:
I thought only the force acting perpendicular to this line serves in this net force equation.
You can either multiply the distance by the force component perpendicular to that line (L*Fsinθ) or multiply the force by the perpendicular distance to the line of action of the force, called the 'moment arm' (Lsinθ*F). Same result either way.
 
Hmm, I think I understand now, thank you very much
 
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