A 1400 kgjet engine is fastened to the fuselage of a passenger jet by just three bolts (this is the usual practice). Assume that each bolt supports one-third of the load. (a) Calculate the force on each bolt as the plane waits in line for clearance to take off. (b) During flight, the plane encounters turbulence, which suddenly imparts an upward vertical acceleration of$\mathbf{2}\mathbf{.}\mathbf{6}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$to the plane. Calculate the force on each bolt now.

Newton’s second law states that the net force on the body is equal to the product of mass of the body and the acceleration with which body is moving.

Apply Newton’s second law to find force on each bolt, when the plane is at rest and when it is in motion.

Formulae:

1. W=mg
2. F=ma

Here,W is weight of the object,m I s mass of the object,g is gravitational acceleration,F is force anda is acceleration of the object.

Apply Newton’s second law, net force on plane is,

$F_{net}=3F-mg$

Here 3F is the force on three bolts.

For rest, $F_{net}=0$

$$\begin{gathered} 0=3F-1400\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2 \\ F=\frac{1400\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2}{3} \\ =4.6\times {10}^3\mathrm{N} \end{gathered}$$

Therefore, the force on each bolt is $4.6\times {10}^3\mathrm{N}$.

When the plane is in the motion, it will have some acceleration. Using the equation of motion, we can write,

$$\begin{gathered} ma=3F-mg \\ F=\frac{m\left(a+g\right)}{3} \\ =\frac{1400\mathrm{kg}\left(2.6/{\mathrm{s}}^2+9.8\mathrm{m}/{\mathrm{s}}^2\right)}{3} \\ =5.8\times {10}^3\mathrm{N} \end{gathered}$$

Therefore, the force on each bolt is $5.8\times {10}^3\mathrm{N}$.