A 6000 kg helicopter is lifting a 1200 kg car with a cable. The helicopter is accelerating upward with 0.350 m/s2. What is the tension in the cable between the car and helicopter? What is the...

A 6000 kg helicopter is lifting a 1200 kg car with a cable. The helicopter is accelerating upward with 0.350 m/s2.

What is the tension in the cable between the car and helicopter?

What is the force of lift from the rotors of the helicopter?

 

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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

Posted on

Hello!

As for many such problems, the significant part is to draw a correct free-body diagram. Please look at the picture attached.

Every force has its cause in a body (or field). Bodies involved in this problem are: the helicopter, the car, the cable, Earth and air. Consider the helicopter and the car first. Denote their masses as `m_1` and `m_2` respectively.

Earth attracts the helicopter with the force `m_1g` and the car with the force `m_2g,` both are directed downwards. The cable pulls the helicopter downwards with the tension force `F_T` and the car with the same force upwards (the cable is assumed to be weightless). By Newton's Third Law `F_T` is also the tension in the cable.

Also, the helicopter presses onto air with the lift force `F_L` downwards, and by the Newton's Third Law air presses onto the helicopter with the same force upwards.

We'll ignore the forces between the other pairs of bodies (although the interaction between the car and air might be significant).

Now recall that Newton's Second Law states that `F=ma,` where `F` is the net force for some object, `m` is its mass and `a` is its acceleration. Write these equations for the helicopter and the car, projected on a vertical axis pointing upwards:

`F_(L)-m_1g-F_(T)=m_1a,`

`F_(T)-m_2g=m_2a.`

The helicopter and the car have the same acceleration `a` because the cable is assumed to be inextensible.

We can found `F_T` from the second equation directly:

`F_(T)=m_2(g+a)=1200*(9.81+0.350) =12192 (N)`

and substitute this into the first equation:

`F_(L)=F_(T)+m_1g+m_1a=m_2(g+a)+m_1(g+a)=(m_1+m_2)(g+a)=`

`=(6000+1200)*(9.81+0.350) =73152 (N).`

 

The answers: the tension in the cable is 12192 N and the force of lift is 73152 N.

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