Answer:
Lifting force, F = 21240 N
Explanation:
It is given that,
Mass of the helicopter, m = 1800 kg
It rises with an upward acceleration of 2 m/s². We need to find the lifting force supplied by its rotating blades. It is given by :
F = mg + ma
Where
mg is its weight
and "ma" is an additional acceleration when it is moving upwards.
So,
F = 21240 N
So, the lifting force supplied by its rotating blades is 21240 N. Hence, this is the required solution.
Answer:
the magnitude of the average contact force exerted on the leg is 3466.98 N
Explanation:
Given the data in the question;
Initial velocity of hand v₀ = 5.25 m/s
final velocity of hand v = 0 m/s
time interval t = 2.65 ms = 0.00265 s
mass of hand m = 1.75 kg
We calculate force on the hand F
using equation for impulse in momentum
F × t = m( v - v₀ )
we substitute
F × 0.00265 = 1.75( 0 - 5.25 )
F × 0.00265 = 1.75( - 5.25 )
F × 0.00265 = -9.1875
F = -9.1875 / 0.00265
F = -3466.98 N
Next we determine force on the leg F
Using Newton's third law of motion
for every action, there is an equal opposite reaction;
so, F = - F
we substitute
F = - ( -3466.98 N )
F = 3466.98 N
Therefore, the magnitude of the average contact force exerted on the leg is 3466.98 N