PHY 2053, Section 3794, Fall 2009, Quiz 10
1. One end of a uniform 7.0 m long rod of weight w is supported
by a cable. The other end rests against the wall, where it is held by
friction. The coefficient of static friction between the wall and the
rod is µs = 0.50. Determine the minimum distance, x, from point A
at which an additional weight w (the same as the weight of the rod)
can be hung without causing the rod to slip at point A.
Let T be the tension in the cable and N the normal force at the wall, with friction given by
FR ≤ µs N . Then:
Fx :
N − T cos(θ) = 0
T sin(θ) + FR − 2 × w = 0, and
L
τ : T × L sin(θ) − w × − w × x = 0,
2
where torque has been calculated around point A. From the first two equations we have:
Fy :
2 × w = T sin(θ) + FR ≤ T sin(θ) + µs × T cos(θ) =⇒ T sin(θ) ≥
2×w
,
1 + µs × cot(θ)
while from the third equation we find
(
T sin(θ) = w ×
)
1 x
+
.
2 L
Putting these together we have
(
)
(
)
2
2
1
1
x≥L
−
= 7.0 ×
−
= 4.9 m.
1 + µs × cot(θ) 2
1 + 0.50 × 1.327 2
2. A planet of radius 8.00 × 103 km spins with an angular velocity of 20.0 × 10−5 rad/s
about an axis through the North Pole. What is the ratio of the normal force experienced
by a person at the equator to that experienced by a person at the North Pole? Assume a
constant gravitational acceleration of 1.60 m/s2 and that both people are stationary relative
to the planet and are at sea level.
At the North Pole we have NN P − mg = 0 =⇒ NN P = mg, while at the equator we have
Neq − mg = −mω 2 R =⇒ Neq = mg − mω 2 R. Thus the ratio Neq /NN P is given by:
Neq
ω2R
(2.00 × 10−4 )2 × 8.00 × 106
=1−
=1−
= 1.000 − 0.200 = 0.800.
NN P
g
1.60