Two blocks each of mass m = 4.10 kg are fastened to the top of an elevator as in the figure below.
(a) If the elevator accelerates upward at 1.6 m/s2, find the tensions T1 and T2 in the upper and lower strings.
T1 ? N
T2 ? N
(b) If the strings can withstand a maximum tension of 104.0 N, what maximum acceleration can the elevator have before the first string breaks?
? m/s2
To solve this problem, we need to analyze the forces acting on the blocks and use Newton's second law of motion.
(a) To find the tensions T1 and T2 in the upper and lower strings, we need to consider the forces acting on each block separately.
For the upper block:
The only external force acting on it is its weight (mg), directed downwards.
The tension T1 in the string pulls the block upwards.
Applying Newton's second law in the vertical direction, we have:
ΣFy = may
T1 - mg = m * (acceleration of the elevator)
For the lower block:
The weight of the lower block (mg) is acting downwards.
The tension T2 in the string pulls the block upwards.
Applying Newton's second law in the vertical direction, we have:
ΣFy = may
T2 - mg = m * (acceleration of the elevator)
Since the two blocks are attached to the same string, the tension in the string is the same for both blocks. Therefore, we can write T1 = T2 = T.
Combining the two equations, we get:
T - mg - mg = m * (acceleration of the elevator)
T - 2mg = m * (acceleration of the elevator)
T = m * (acceleration of the elevator) + 2mg
Now we can substitute the given values:
m = 4.10 kg
acceleration of the elevator = 1.6 m/s^2
g = acceleration due to gravity = 9.8 m/s^2
T = (4.10 kg * 1.6 m/s^2) + 2(4.10 kg * 9.8 m/s^2)
T = 6.56 N + 80.36 N
T ≈ 86.92 N
Therefore, the tensions in both strings are approximately T1 = 86.92 N and T2 = 86.92 N.
(b) To find the maximum acceleration the elevator can have before the first string breaks, we need to consider the maximum tension the strings can withstand.
From the previous calculation, we know that the tension in the string is T = 86.92 N.
To find the maximum acceleration, we can again use the equation:
T = m * (acceleration of the elevator) + 2mg
Rearranging the equation, we have:
(acceleration of the elevator) = (T - 2mg) / m
Substituting the given values:
T = 104.0 N (maximum tension)
m = 4.10 kg
g = 9.8 m/s^2
(acceleration of the elevator) = (104.0 N - 2(4.10 kg * 9.8 m/s^2)) / 4.10 kg
(acceleration of the elevator) = (104.0 N - 80.36 N) / 4.10 kg
(acceleration of the elevator) ≈ 5.76 m/s^2
Therefore, the maximum acceleration the elevator can have before the first string breaks is approximately 5.76 m/s^2.
To solve this problem, we can break it down into two separate parts: one for finding the tensions in the strings and another for finding the maximum acceleration before the first string breaks.
(a) Finding the tensions T1 and T2:
From the given information, we have two blocks of mass m = 4.10 kg each. Let's assume the acceleration of the elevator is a = 1.6 m/s^2. Using Newton's second law, we can find the force acting on each block:
F = ma
For the upper block:
F1 = m * a = 4.10 kg * 1.6 m/s^2 = 6.56 N
The force acting on the upper block is the sum of the tension T1 pulling upward and the force due to its weight:
F1 = T1 - m * g
where g is the acceleration due to gravity (9.8 m/s^2). Substituting the known values:
6.56 N = T1 - 4.10 kg * 9.8 m/s^2
T1 = 6.56 N + 4.10 kg * 9.8 m/s^2
T1 = 6.56 N + 40.18 N
T1 = 46.74 N
For the lower block:
The force acting on the lower block is the sum of the tension T2 pulling downward and the force due to its weight:
F2 = T2 + m * g
Since the lower block is connected to the upper block, the tension in the lower string T2 is equal to the tension in the upper string T1:
F2 = T2 + m * g = T1
Substituting the known values:
T2 + 4.10 kg * 9.8 m/s^2 = 46.74 N
T2 + 40.18 N = 46.74 N
T2 = 46.74 N - 40.18 N
T2 = 6.56 N
Therefore, the tensions T1 and T2 in the upper and lower strings are:
T1 = 46.74 N
T2 = 6.56 N
(b) Finding the maximum acceleration before the first string breaks:
The maximum tension the strings can withstand is given as 104.0 N. This means that if the tension exceeds 104.0 N, the string will break. Since T1 is the larger tension, we need to find the maximum acceleration that would result in T1 being equal to or less than 104.0 N.
Using the equation for T1 calculated earlier:
T1 = 46.74 N
We can set T1 equal to the maximum tension and solve for the maximum acceleration:
104.0 N = 6.56 N + 4.10 kg * a
Solving for a:
4.10 kg * a = 104.0 N - 6.56 N
4.10 kg * a = 97.44 N
a = 97.44 N / 4.10 kg
a ≈ 23.78 m/s^2
Therefore, the maximum acceleration the elevator can have before the first string breaks is approximately 23.78 m/s^2.