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One force acting on a machine part is
F = (-5.00N)i + (4.00N)j.
The vector from the origin to the pouint where the force is applied is
r= (-4.50m)i + (1.50m)j.


a) Determine the direction of the torque using the right hand rule.
b) Calculate the magnitude of the torque produced by this force.
c) From the answer in b), what is the direction of the torque? (this must be in match with the answer in b))

2007-08-25 21:43:26 · 1 answers · asked by nadanilnone 1 in Science & Mathematics Physics

1 answers

The center of rotation of the part is at the origin. This is implied by the statement of the problem. Fortunately we can handle the vertical and the horizontal components of the force separately, and then combine them. We simply need to keep track of the signs of the torques using the right hand rule.
Label the positive x-axis i, and the positive y axis j, and plot the point (-4.5, 1.5). this is the point at which the force is applied. Place your right hand over the origin, thumb pointing up and fingers curled. Your fingers depict a counter-clockwise direction. notice that the i component of the force, -5.0N, applied at a moment arm of +1.5m applies a positive torsional moment at the origin. Also notice that the j component of the force, 4.00N applied at a moment arm of -4.5m applies a negative torsional moment at the origin.
Here's the calculation:
5.00(1.5) - 4(4.5) = -10.5
a) Negative
b) -10.5N
c) Negative(clockwise)

2007-08-26 00:43:12 · answer #1 · answered by jsardi56 7 · 3 5

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