Recall the experiment activities. If the meterstick is pivoted at the 20cm mark,
ID: 1610098 • Letter: R
Question
Recall the experiment activities. If the meterstick is pivoted at the 20cm mark, a force of 3.48N is applied at 63cm mark, what is the maximum possible torque (in unit of Nm) by the force? When several forces are applied on a hinged rod, the greatest force will have the greatest torque. True False Two torques are applied on a pivoted rod. One is a clockwise torque of magnitude 18.6N middot m, the other is a counterclockwise torque of magnitude 16N middot m. What is the total torque on the rod (in unit of N middot m)? (Don't forget to include proper sign to indicate whether the total torque is clockwise or counterclockwise) 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?Explanation / Answer
(1)
formula torque is
T = rF
= ( 0.63 m - 0.2 m) 3.48 N
=1.49 N m
(b)
rue because torque is proportional to F
(c)
T net = TC- TCCW = 18.6-16 = 2.6 N m
(d)
neet figure for question (7)
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