June 13, 2008
PHY2053 Discussion
Quiz 4 (Chapter 4)
Name: UFID:
*1. (5pts) Two telephone poles are separated by a distance of 30.0 m. A bird with a weight of 25.0 N rests at the center of the telephone line between the two poles. The wire sags 0.300 m. What is the tension in the string?
To resolve the tension in components, we need the angle the telephone line makes to the horizontal. Since the horizontal distance between the bird and the telephone pole is 15 m, we have tanθ = 0.3/15 ⇒ θ = tan -1 (0.3/15) = 1.15˚.
We apply the equilibrium condition to horizontal and vertical directions, and solve the equations simultaneously:
T
R cosθ-T
L cosθ = 0 ⇒ T
R
= T
L
= T
T
R sinθ+T
L sinθ-W = 0 ⇒ 2Tsinθ = W ⇒ T = W/(2sinθ) = 25/(2sin1.15˚) = 623 N
*2. (5pts) Two blocks are suspended from the ceiling of an elevator as follows: A rope is attached to the ceiling and connected to 10.0-kg block. Another rope is attached to the lower surface of the 10-kg block and then connected to a 15.0-kg block. The elevator accelerates downward at 3.00 m/s. Find the tension in the top rope.
We take +x-axis upward, then the acceleration of both blocks is -3 m/s². Applying the
Newton’s 2 nd law to the bottom block, we have m
1 a = T
1
-m
1 g ⇒ T
1
= m
1
(g+a) = 15 × (9.8-3) = 102 N
Applying the Newton’s 2 nd law to the top block, we have m
2 a = T
2
-T
1
-m
2 g ⇒ T
2
= T
1
+m
2
(g+a) = 102+10 × (9.8-3) = 170 N

**3. (5pts) A 10.0 kg block is on 45.0˚ incline. The block is attached to a string that passes over a pulley at the top of the incline and then connected to a 4.00 kg hanging object. The system starts from rest and the block slides 0.800 m down the incline in 2.50 s. Find the coefficient of kinetic friction between the block and the incline.
The acceleration of the block is given by one of the kinematics equations:
Δx = (1/2)at² ⇒ a = 2Δx/t² = 2 × 0.8/2.5² = 0.256 m/s².
Taking +x axes to the directions of motions, we apply Newton’s 2 nd law to the hanging object and the block and solve them simultaneously. m
1 a = T-m
1 g ⇒ T = m
1
(a+g) = 40.2 N m
2 a = m
2 gsinθ – μ k m
2 gcosθ-T ⇒ μ k
= (m
2 gsinθ-m
2 a-T)/(m
2 gcosθ) = 0.383
***4. (5pts) A 5.00-kg block rides on a 10.0-kg block. They are on the horizontal ground.
A string is attached to the top block and a person pulls the string at an angle of 15.0˚ to the horizontal. The coefficient of static friction between the two blocks is 0.500 and the coefficient of kinetic friction between the bottom block and the ground is 0.100.
Assuming the person is initially moving the two blocks together, find the maximum force that can be exerted on the string without causing the top block to slip.
Solving the equilibrium condition in the vertical direction, the normal forces on the top block and the bottom block are n+Tsinθ-mg = 0 ⇒ n = mg-Tsinθ
N-n-Mg = 0 ⇒ N = Mg+n = (M+m)g-Tsinθ
When the top block is on the verge of slipping, the static friction on the top block is maximum and the two blocks still move together. Applying Newton’s 2 nd law in the horizontal directions, we have ma = Tcosθ-μ s n = T(cosθ+μ s sinθ)-μ s mg
Ma = f s
-f k
= μ s
(mg-Tsinθ)-μ k
((M+m)g-Tsinθ) = (μ s
-μ k
)mg-μ k
Mg-(μ s
-μ k
)Tsinθ
⇒ mMa = T(cosθ+μ s sinθ)M-μ s mMg = (μ s
-μ k
)m²g-μ km
Mg-(μ s
-μ k
)mTsinθ
⇒ T = (μ s
-μ k
)m(m+M)g/[Mcosθ+(μ s
M+(μ s
-μ k
)m)sinθ]
= 0.4
× 5 × 15 × 9.8/(10 × cos15˚+7 × sin15˚) = 25.6 N