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.

(b) If the strings can withstand a maximum tension of 96.0 N, what maximum acceleration can the elevator have before the first string breaks?

I will be happy to critique your thinking.

Consider, T1 - T2 - mg = ma

Also, T2 - mg = ma

Then,

T1 = 2ma + 2mg
&
T2= mg + ma

so,

T1 = 93.48 N
&
T2 = 46.74 N

Now to find max acc using given Tension Force 96. Set equation 1 using T1 as 96 and now find a.
your answer should equal a = 1.90 m/s2

Enjoyyy...

To find the tensions T1 and T2, we can begin by analyzing the forces acting on each block.

(a)

Considering the first block:

1. The weight of the block acts downwards and is given by W1 = m * g, where m is the mass of the block and g is the acceleration due to gravity (approximately 9.8 m/s^2).
2. The tension force T1 acts upwards on the block, pulling it.

Applying Newton's second law to the first block, we have:

ΣF1 = m * a1

where ΣF1 is the sum of the forces acting on the first block, m is the mass of the block, and a1 is the acceleration of the first block.

The forces acting on the first block are:

ΣF1 = T1 - W1

Substituting the expression for weight, we have:

T1 - W1 = m * a1

Now, let's consider the second block:

1. The weight of the second block acts downwards and is given by W2 = m * g.
2. The tension force T2 acts upwards on the second block, pulling it.

Again, applying Newton's second law to the second block, we have:

ΣF2 = m * a2

where ΣF2 is the sum of the forces acting on the second block, m is the mass of the block, and a2 is the acceleration of the second block.

The forces acting on the second block are:

ΣF2 = T2 - W2

Substituting the expression for weight, we have:

T2 - W2 = m * a2

Given that the elevator accelerates upwards at 1.6 m/s^2, we can substitute this value for a1 and a2 in the above equations:

T1 - (m * g) = m * a1

T2 - (m * g) = m * a2

Now, we can substitute the given values for mass (m = 4.10 kg), acceleration due to gravity (g ≈ 9.8 m/s^2), and elevator acceleration (a = 1.6 m/s^2) to solve for T1 and T2.

(b)

To find the maximum acceleration the elevator can have before the first string breaks, we need to determine the maximum tension force the string can withstand.

Given that the maximum tension the strings can withstand is 96.0 N, we can set the tensions T1 and T2 equal to this maximum value and solve for the maximum acceleration.

Using the previously derived expressions for T1 and T2, we set both tensions equal to 96.0 N:

T1 - (m * g) = 96.0
T2 - (m * g) = 96.0

Considering the limiting case when T1 = 96.0 N, we can solve for a1:

96.0 - (m * g) = m * a1

Likewise, for the limiting case when T2 = 96.0 N, we solve for a2:

96.0 - (m * g) = m * a2

Therefore, the maximum acceleration the elevator can have before the first string breaks is equal to a1 or a2, whichever is smaller.