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Chapter 7: Q. 44 (page 180)

A 3200kghelicopter is flying horizontally. A 250kgcrate is suspended from the helicopter by a massless cable that is constantly 20°from vertical. What propulsion force F u prop is being provided by the helicopter’s rotor? Air resistance can be ignored. Give your answer in component form in a coordinate system where i^points in the direction of motion and j^points upward.

Short Answer

Expert verified

Propulsion force,F→=12.3kNi^+33.8kNj^

Step by step solution

01

Given information 

The mass of the helicopter M=3200kg

The mass of the cratem=250kg

02

Explanation 

The expression to calculate the vertical component of force is,

fy=mg

The cable makes an angle with the vertical component of the force.

tanθ=fxfyfx=mgtanθ

The acceleration of both the systems remains the same so, the horizontal component of the force on the system is given as,

localid="1649168708556" fxm=fyM+mmgtanθm=FxM+mFx=M+mgtanθFx=3200kg+250kg9.8m/s2tan20°=12305.8N

Now calculate vertical force using

Fy=M+mg=3200kg+250kg9.8m/s2=33.8kN

Therefore the propulsion force provided by the helicopter's rotor is

localid="1649169065712" F→=12.3kNi^+33.8kNj^

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