a. 0.667 m b. 0.770 m c. 0.834 m d. 0.725 m e. 0.783 m A meter stick is placed f

Question

a. 0.667 m
b. 0.770 m
c. 0.834 m
d. 0.725 m
e. 0.783 m

A meter stick is placed flat on a table, and one end of themeter stick is attached to the table. The other end of the meterstick can thus rotate freely around this attached end (the axis)over the surface of the table. Two forces, both parallel to thetabletop, are applied to the stick in such a way that the nettorque is zero. One force has a magnitude of 4.00 N and is appliedperpendicular to the stick at the free end. The other force isapplied on the opposite side of the above perpendicular force, butit has a magnitude of 6.00 N and acts at a 60.0 degree angle withrespect to the stick (see diagram). Where along the stick is the6.00 N force applied? Express this distance with respect to theaxis of rotation. a. 0.667 m b. 0.770 m c. 0.834 m d. 0.725 m e. 0.783 m

Explanation / Answer

Since the net torque is zero, the addition of both torquesmust equal 0.
The 4N force pushes up and the torque will be in the positivecounter-clockwise direction while the other force is directeddownward and the torque will be in the negative clockwisedirection. Also you must take sin 60 * the Force since you only want they-component of the force since that is the component that iscausing the bar to spin. Net = 1 + 2 =0Nm F1r1 + F2r2 =0Nm F1r1 =F2r2 4N * 1m = (sin 60)6N * r2 r2 = 0.77m r2 = 0.77m

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