Solveeit Logo
Login

Question

Question: Helicopter of mass \[1000kg\] Rises with a vertical acceleration of \[15m/{{\sec }^{2}}\] . The crew...

Helicopter of mass 1000kg1000kg Rises with a vertical acceleration of 15m/sec⁡215m/{{\sec }^{2}} . The crew and the passengers weigh 300kg300kg. Give the magnitude and direction of the (A) Force on the Floor by the crew and passengers. (B) Action of the rotor of the helicopter on the surrounding air. (C)Force on the helicopter due to the surrounding air.

Explanation

Solution

Use the formula of Newton’s second law of motion.
Use the concept of balancing the force acted.
Use the gravitational acceleration in all calculations.

Complete step by step solution:
The second law of motion describes what happens to the massive body when acted upon by an external force. The 2nd law of motion states that the force acting on the body is equal to the product of its mass and acceleration.

Newton’s second law of motion states that,
The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.
In this question the helicopter rises vertically then we have to face on the floor in different conditions. So reaction force is calculated by making F.B.D. and using the Law of Motion.

Man of helicopter mH=1000kg{{m}_{H}}=1000kg
Total man of crew and passengers m=300kg{{m}_{{}}}=300kg
Acceleration of helicopter a=15m/s2a=15m/{{s}^{2}}
Total man of system mT=1000+300{{m}_{T}}=1000+300
=1300kg=1300kg

A. Reaction force applied by the crew and passengers on the floor
R−mg=maR-mg=ma
R=mg+maR=mg+ma
=m(g+a)=m(g+a)
=300(10+15)=7500 N=300(10+15)=7500\ N
So the face exerted by the passengers and crew members is 7500 N

B. Let action of rotor on the surrounding air by the helicopter R’
Here we have to consider total man. So
R1−mTg=mTa{{R}^{1}}-{{m}_{T}}g={{m}_{T}}a
R1−mT(g+a){{R}^{1}}-{{m}_{T}}(g+a)
=1300(10+15)=32500  N=1300(10+15)=32500\ \ N
So, the reaction face of the rotor on the surrounding air will be 32500 N

C. The face on the helicopter due to surrounding air will be the reaction of the face applied by the rotor on the air.
So it is same 32500N.32500N.

Note: 1. Students should have clear knowledge of force.
2. Students also have complete knowledge of law of motion and free body diagrams.
3. Use the balancing of force very carefully; causes vary the value of the final result.