For the meter stick shown above, the force F_1 = 10.0 N acts at 10.0 cm. What is
ID: 1550869 • Letter: F
Question
For the meter stick shown above, the force F_1 = 10.0 N acts at 10.0 cm. What is the magnitude of the torque due to Fi about an axis through point A perpendicular to the page? Is the torque clockwise or is counterclockwise? _____ In the figure the force F_2 = 15.0 N acts at the point 70.0 cm. What is the magnitude of the torque due to Fr about an axis through point B and perpendicular to the page? Is the torque clockwise or is it counterclockwise? _____ In the figure above, if the mass m_1 = 0.100 kg acts at 20.0 cm, what is the value of mass m2 that must be placed at the position 70.0 cm shown to put the system in equilibrium? Assume the meter stick is uniform and symmetric. Show your work. _____Explanation / Answer
1)
Torque = F* d_perp
F = Force
d_perp = perpendicular distance
Torque = (10.0) (0.10) = 1 Nm
Clockwise
2)
Torque = F d_perp
= (15.0) ( 0.20)
= 3 N-m
Clockwise
3)
m1 gd1 = m2gd2
=> (0.100)(9.80)(0.300) = m2 (9.80)(0.200)
m2 = (0.100)(0.300)/(0.200) = 0.150 kg