Recall the experiment activities. If the meterstick is pivoted at the 20cm mark,
ID: 1998629 • Letter: R
Question
Recall the experiment activities. If the meterstick is pivoted at the 20cm mark, a force of 3.1 N is applied at 82cm mark, what is the maximum possible torque fin unit of Nm) by the force? A meterstick has a mass 75grams. and its center of mass is at 49.3cm. If the meterstick is pivoted at the 5cm mark, and a force is applied at the 95cm mark as shown below, how much must the force be (in unit of newton) to keep the meterstick in equilibrium? Recall the experiment activities. If the meterstick is pivoted at the 20cm mark, a force of 2N is applied at 56cm mark, what is the maximum possible torque fin unit of Nm) by the force? Two torques are applied on a pivoted rod. One is a clockwise torque of magnitude 37.1N middot m, the other is a counterclockwise torque of magnitude 30.7N middot m. What is the total torque on the rod (in unit of N m)? (Don't forget to include proper sign to indicate whether the total torque is clockwise or counterclockwise)Explanation / Answer
here,
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Pivoted point , x = 0.2 m
force applied at x' = 0.82 m
the maximum possible torque by the force , T = F *( x' - x)
T = 3.1 * ( 0.82 - 0.2)
T = 1.9 N.m
the maximum possible torque by the force is 1.9 N.m
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