# Boxes A and B are attached by a rope. They are on a platform floor attached to a helicopter that is falling. Box A is pulled 35^@ to the surface. What is the tension between A and B as they slide horizontally on the platform?

## Box A ($12.0 k g$) and B ($7.0 k g$) are attached by a rope. They are on a platform floor attached to a helicopter that is falling at a rate of $2.5 \frac{m}{s} ^ 2$. Box A is pulled with a force of $42 N$, ${35}^{\circ}$ to the surface. The coefficient of friction for both A and B against the floor is ${\mu}_{k} = 0.20$. What is the tension between A and B as they slide horizontally on the platform?

Jun 25, 2016

$T \approx 14.46 N$

#### Explanation:

As the Helicopter is falling with acceleration $2.5 \frac{m}{s} ^ 2$, taking normal acceleration due to gravity as $10 \frac{m}{s} ^ 2$ we can say that under the reference frame of accelerated helicopter the effective value of acceleration due to gravity within falling helicopter is ${g}_{e} = \left(10 - 2.5\right) \frac{m}{s} ^ 2 = 7.5 \frac{m}{s} ^ 2$

Given

• ${m}_{A} \to \text{Mass of box A} = 12 k g$

• ${m}_{B} \to \text{Mass of box B} = 7 k g$

• ${\mu}_{k} \to \text{Coefficient of kinetic friction } = 0.2$

• $\text{Applied force on A"=42N ," acting " 35^@ "with horizontal}$

• $\text{Horizontal component of 42N} = 42 \cos 35 N$

• $\text{Vertictal component of 42N} = 42 \sin 35 N$

Let the combined system is moving with acceleration $a \frac{m}{s} ^ 2$ when pulled by 42N force as shown in the figure and the tension between A and B is T

Then considering forces acting on 7 kg box (B ) we can write

$T - {\mu}_{k} \times {m}_{B} \times {g}_{e} = 7 a$

$\implies T - 0.2 \times 7 \times 7.5 = 7 a$

Dividing both sides by 7

$\implies \frac{T}{7} - 0.2 \times 7.5 = a \ldots \ldots . . \left(1\right)$

Again considering forces acting on12kg box (A ) we can write

$42 \cos 35 - {\mu}_{k} \left({m}_{A} \times {g}_{e} - 42 \sin 35\right) - T = 12 a$

$\implies 42 \cos 35 - 0.2 \left(12 \times 7.5 - 42 \sin 35\right) - T = 12 a$

Dividing both sides by 12 we get

$\frac{42 \cos 35}{12} - \frac{0.2 \left(12 \times 7.5 - 42 \sin 35\right)}{12} - \frac{T}{12} = a \ldots . . \left(2\right)$

Now subtracting (2) from (1) we get

$\frac{T}{7} - 0.2 \times 7.5 - \frac{42 \cos 35}{12} + \frac{0.2 \left(12 \times 7.5 - 42 \sin 35\right)}{12} + \frac{T}{12} = 0$

$\implies \frac{19}{84} T = 0.2 \times 7.5 + \frac{42 \cos 35}{12} - \frac{0.2 \left(12 \times 7.5 - 42 \sin 35\right)}{12}$

$\implies \frac{19}{84} T = \cancel{0.2 \times 7.5} + 2.87 - \cancel{0.2 \times 7.5} + \frac{0.2 \times 42 \sin 35}{12}$

$\implies \frac{19}{84} T = 2.87 + 0.7 \sin 35 = 3.27$

$T = \frac{84 \times 3.27}{19} N = 14.46 N$