91Ó°ÊÓ

Frictional force acts when there is relative motion between two bodies or the possibility of relative motion arises.

The direction of the frictional force, in this case, is such that it tries to stop that relative motion.

The mass of the helicopter,$M=7180\mathrm{kg}$

The mass of the frame,$m=1080\mathrm{kg}$

The acceleration of the system,$a=0.80\mathrm{m}/{\mathrm{s}}^2$

Let the tension in the cable connecting the helicopter to the frame be ${\stackrel{→}{F}}_{\mathrm{T}}$, and the lift force be $F_{\mathrm{lift}}$.

Applying Newton’s second law on the system of the helicopter and frame in the vertical direction, you get:

$F_{\mathrm{lift}}-\left(Mg+mg\right)=\left(m+M\right)a$

Substituting the values in the above equation, you get:

$\begin{array}{c}{F}_{\mathrm{lift}}=\left(7180+1080\right)\left(9.8+0.80\right)\\ =87556\mathrm{N}\end{array}$

Thus, the magnitude of the lift force is 87556 N.

The tension force in the cable can be found out by analysing the motion of the frame. Given below is the FBD of the frame.

Applying Newton’s second law in the vertical direction, you get:

$\left|{\stackrel{→}{F}}_{\mathrm{T}}\right|-mg=ma$

Substituting the values, you get:

$\left|{\stackrel{→}{F}}_{\mathrm{T}}\right|-1080×9.8=1080×0.80$

Solving the above equation, you get:

$\left|{\stackrel{→}{F}}_{\mathrm{T}}\right|=11448\mathrm{N}$

Thus, the tension force in the cable will be 11448 N.

As the cable is massless, the tension force in the cable will be the same throughout its entire length. Thus, the force exerted by the cable on the helicopter will be tension force.

$\left|{\stackrel{→}{F}}_{\mathrm{T}}\right|=11448\mathrm{N}$