Solution.
We write StA = 0 for the system
Wd + Mg(L/2) - FTLsinq = 0;
(15N)(3m) + (5kg)(9.80m/s2)(10m/2) - FT(10m)sin60° = 0,
which gives FT = 33.5N.
QUIZ 9
| 1. A 10-m-long uniform rod of mass M = 5kg is
pivoted at one end A and
supported at the other end by a rope as shown in the figure. A 15N weight
hangs at 3m from A. Find the tension FT in the rope.
Solution. We write StA = 0 for the system Wd + Mg(L/2) - FTLsinq = 0; (15N)(3m) + (5kg)(9.80m/s2)(10m/2) - FT(10m)sin60° = 0, which gives FT = 33.5N. |
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| 2. The right
car is easy to turn over. As shown in the figure, when the cars
tilt the same angle, the right wheels' edges serve as the pivot point, the
right car's CM is out of the base of the car. The torque due to its
weight causes the car turn clockwise and the car turns over. The left
car's CM is within its base when tilts. The torque due to its weight will
turn the car counterclockwise, the car return to its original position.
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