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?

1 Answer
Jun 25, 2016

Answer:

#T~~14.46N#

Explanation:

self drawn

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#