Question

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 (`12.0kg`) and B (`7.0kg`) are attached by a rope. They are on a platform floor attached to a helicopter that is falling at a rate of `2.5m/s^2`. Box A is pulled with a force of `42N`, `35^@` 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?

Answer 1

`T~~14.46N`

Explanation:

As the Helicopter is falling with acceleration `2.5m/s^2`, taking normal acceleration due to gravity as `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)m/s^2=7.5m/s^2`

Given

• `m_A->"Mass of box A" =12kg`

• `m_B->"Mass of box B" =7kg`

• `mu_k->"Coefficient of kinetic friction " =0.2`

• `"Applied force on A"=42N ," acting " 35^@ "with horizontal"`

• `"Horizontal component of 42N"=42cos35N`

• `"Vertictal component of 42N"=42sin35N`

Let the combined system is moving with acceleration `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-mu_kxxm_Bxxg_e= 7a`

`=>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`