Question

Rope A is tied to block 1, and rope B is attached to both block 1 and block 2 as shown in the diagram. Block 1 has a mass of 4.2 kg and block 2 has a mass of 2.6 kg. You lift both blocks straight up. Calculate the magnitude of tension in each of the ropes when the blocks

Move at constant velocity of 1.5 m/s [up]
Find the magnitude of tension in each rope when the blocks are accelerating at 1.2 m/s^2 [up].
The maximum tension the strings can withstand is 90. N. Knowing this, determine the maximum acceleration of the blocks that would not break the rope.

Answer 1

Hi there!

Part 1:

If the blocks are moving at a constant velocity:

∑F = 0

Begin by summing the forces acting on each block. Let the upward direction be positive.

∑F₁ = Ta - M₁g - Tb

∑F₂ = Tb - M₂g

Sum the forces:

∑F = Ta - M₁g - Tb + Tb - M₂g

∑F = Ta - M₁g - M₂g = 0

Solve for Tension A:

Ta = M₁g + M₂g (Let g = 9.8 m/s²)

Ta = 4.2(9.8) + 2.6(9.8) = 66.64 N

Now, solve for tension B using the summation of ∑F₁:

0 = Tb - M₂g

Tb = (2.6* 9.8) = 25.48 N

Part 2:

We can use the same method, but incorporate the acceleration:

∑F = Ta - M₁g - M₂g

(M₁ + M₂)a = Ta - M₁g - M₂g

(M₁ + M₂)a + M₁g + M₂g = Ta

(4.2 + 2.6)(1.2) + 4.2(9.8) + 2.6(9.8) = 74.8 N

∑F₂ = Tb - M₂g

M₂a + M₂g = Tb  = 28.6 N

Part 3:

Since the top string experiences most of the tension, we can use its equation to calculate the maximum acceleration:

∑F = Ta - M₁g - M₂g

(M₁ + M₂)a = Ta - M₁g - M₂g

a = (90 - M₁g - M₂g)/(M₁ + M₂)

a = 3.435 m/s²