Three forces are applied to a wheel of radius 0.350 m. One force is perpendicular to the rim, one is tangent to it, and the other one makes a 40 degree angle with the radius. What is the torque that is net the wheel considering these three forces for an axis perpendicular toward wheel and moving through it is center? (The force perpendicular toward rim = 11.9 N; the potent force making a 40 level angle = 14.6N; The force which tanget to it = 8.50 N)

You will need to determine the tangential force that is net. The force that is perpendicular have zero force across the tangent. The force that is tangential of course, would be 100% tangential. The third force would be scaled according to the cosine of the angle between it and the radius between it and the tangent line, or the sine of the angle:

14.6*sin(radians(40)) = 9.385N

Now, the web torque will depend on the WAY of this two forces which have a result. They add up: 8.50+9.385 if they are both in the same direction, say clockwise = 17.885N, but they subtract: 8.50-9.385 if they are opposed = -0.885N, which, 0.885N in way opposite the force that is tangential.

Finally, to determine the torque that is net we multiply the internet tangential force using the radius of this wheel:

17.885 * 0.350 = 6.26 Nm

or…

-0.885 * 0.350 = -0.310 Nm

It must be 0.31, to the web page.

tnx the concept.

no clue