Why you are calculating the the torque due to gravity, why do you multiply by co

Question

Why you are calculating the the torque due to gravity, why do you multiply by cos@ instead of sin@? If you can draw the vector diagram of the mg force and points out to me which component causes the torque. It woud be awesome!

Let theta be the angle the plane of the loop makes with the horizontal as shown in the sketch at the right. Then, the angle it makes with the vertical is theta = 90.0degree - theta. The number of turns on the loop is N = L / circumference = 4.00/4(0.100 m) = 10.0 The torque about the z axis due to gravity is , where s = 0.100 m is the length of one side of the loop. This torque tends to rotate the loop clockwise. The torque due to the magnetic force tends to rotate the loop counterclockwise about the z axis and has magnitude At equilibrium, This reduces to tan theta = mg/2NBIs =(0.100 kg)(9.80 m/s2)/2(10.0)(0.0100 T)(3.40 A)(0.100 m) = 14.4 Since tan theta = tan(90.0degree-phi) = cot theta, the angle the loop makes with the vertical at equilibrium is phi = cot-1 (14.4) = 3.97degree .

Explanation / Answer

here torque is calculated about z-axiz , so we are required to knoe the perpendicular distance of force application from axis

as s/2 is the length along the object, for finding perdiculardistance from z-axis, its component is taken on perpendicular line by multiplying by cos (theta) as per given in the figure

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