Summary







































pse6 12.24
Two identical, uniform bricks of length L are stacked as shown. What is the maximum x possible before the bricks topple and fall?
A. L
B. 3L/4
C. L/2
D. L/3

Answer



















Walker5e Ex 11-11
An 85-kg person stands on a lightweight ladder, as shown in the figure. The floor is rough; hence, it exerts both a normal force f1 and a frictional force f2 on the ladder. The wall is frictionless and exerts only normal force f3. Use the dimensions in the figure to find the magnitude of force f2.
person on a ladder in equilibrium A. 85 N
B. 153 N
C. 680 N
D. 833 N
Answer



















gc6 9.36
The Leaning Tower of Pisa is 55 m tall and about 7.0 m in diameter. The top is 4.5 m off center. How much farther can it lean before it becomes unstable?
A. 1.0 m
B. 2.5 m
C. 3.5 m
D. 4.5 m
Answer



















Walker4e
Which portion of the sphere of mass M exerts the greatest force on the small mass m? Walker4e figure 12-3
A. portion A
B. either portion D or C
C. portion B
D. All portions exert the same force on m.
Answer



















klm
Suppose you and the person next to you are each 70 kg and are sitting 0.50 m apart. What is the force of attraction between you and they?
A. 9.34×10−9 N
B. 1.87×10−8 N
C. 6.54×10−7 N
D. 1.31×10−6 N
Answer



















klm
Two spacecraft in outer space attract each other with a force of 1×106 N. What would the attractive force be if they were twice as far apart?
A. 4.0×106 N
B. 2.0×106 N
C. 5.0×105 N
D. 2.5×105 N
Answer



















PoP5 11.1mod
What is the horizontal gravitational acceleration experienced by adjacent 40,000 metric ton ships that are 25 m apart?
A. 0.00427 m/s2
B. 4.27 µm/s2
C. 427 nm/s2
D. 0.427 pm/s2
Answer






















pop4 11.8
What is gEarth at the point where gEarth = gMoon? The centers of the Earth and Moon are 384,400 km apart, and the masses are MEarth = 5.97×1024 kg and MMoon = 7.35×1022 kg.
A. 2.72 mm/s2
B. 3.33 mm/s2
C. 115 mm/s2
D. 3.46 m/s2
Answer










































answer


B. 3L/4
The center of mass of the top brick must be above the point of support, and the center of mass of the two-brick combination must also be above its point of support. The solution becomes more interesting as you consider more bricks:

solution equation first part
solution equation second part
solution equation third part


If you're interested, read more about how alternative strategies can lead to a maximum overhang of N1/3



















answer
person on a ladder in equilibrium
B. 153 N
Set the torques about the bottom of the ladder (marked "Axis") equal to zero. Examine the diagram to see the torque due to the person's weight is −mgb (clockwise, so negative), and the torque due to f3 is +a f3. You will find f3 is 153 N, then set the sum of the horizontal forces equal to zero to find f2 = f3.
























answer

B. 2.5 m
The top would have to move one diameter in order to put the center gravity, assumed to be halfway down, over the pivot point of one side of the base. It can therefore move 7.0 m − 4.5 m = 2.5 m further.
























answer

Walker4e figure 12-3
A. portion A

The force increases when the distance r decreases, so the nearest portion of the mass M exerts the greatest force on m. The combined force from all portions acts as if all of the mass M were concentrated at its center.




















answer

D. 1.31×10−6 N
Picture (893x144, 4.9Kb)



















answer

D. 2.5×105 N
Picture (524x282, 4.9Kb)



















answer

B. 4.27 µm/s2
solution equation



















answer

B. 3.33 mm/s2
solution equation