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 $35^{\circ}$ $35^@$ to the surface. What is the tension between A and B as they slide horizontally on the platform?

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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 $35^{\circ}$ $35^@$ to the surface. What is the tension between A and B as they slide horizontally on the platform?

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

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Answer 1 · 2016-06-25T04:33:16

$T \approx 14.46 N$ $T~~14.46N$

Explanation:

self drawn

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

Given

$m_{A} \to Mass of box A = 12 k g$ $m_A->"Mass of box A" =12kg$
$m_{B} \to Mass of box B = 7 k g$ $m_B->"Mass of box B" =7kg$
$μ_{k} \to Coefficient of kinetic friction = 0.2$ $mu_k->"Coefficient of kinetic friction " =0.2$
$Applied force on A = 42 N, acting 35^{\circ} with horizontal$ $"Applied force on A"=42N ," acting " 35^@ "with horizontal"$
$Horizontal component of 42N = 42 cos 35 N$ $"Horizontal component of 42N"=42cos35N$
$Vertictal component of 42N = 42 sin 35 N$ $"Vertictal component of 42N"=42sin35N$

Let the combined system is moving with acceleration $a \frac{m}{s^{2}}$ $am/s^2$ 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

$T - μ_{k} \times m_{B} \times g_{e} = 7 a$ $T-mu_kxxm_Bxxg_e= 7a$

$\Rightarrow T - 0.2 \times 7 \times 7.5 = 7 a$ $=>T-0.2xx7xx7.5= 7a$

Dividing both sides by 7

$=>T/7-0.2xx7.5= a ........(1)$

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

$42cos35-mu_k(m_Axxg_e-42sin35)-T=12a$

$=>42cos35-0.2(12xx7.5-42sin35)-T=12a$

Dividing both sides by 12 we get

$(42cos35)/12-(0.2(12xx7.5-42sin35))/12-T/12=a.....(2)$

Now subtracting (2) from (1) we get

$T/7-0.2xx7.5-(42cos35)/12+(0.2(12xx7.5-42sin35))/12+T/12=0$

$=>19/84T=0.2xx7.5+(42cos35)/12-(0.2(12xx7.5-42sin35))/12$

$=>19/84T=cancel(0.2xx7.5)+2.87-cancel(0.2xx7.5)+(0.2xx42sin35)/12$

$=>19/84T=2.87+0.7sin35=3.27$

$T=(84xx3.27)/19N=14.46N$

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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 $35^{\circ}$ $35^@$ to the surface. What is the tension between A and B as they slide horizontally on the platform?

1 Answer

Explanation:

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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^@35∘35^@ to the surface. What is the tension between A and B as they slide horizontally on the platform?

1 Answer

Explanation:

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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 $35^{\circ}$ $35^@$ to the surface. What is the tension between A and B as they slide horizontally on the platform?