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Chapter 4: Problem 89

A helicopter of mass \(M\) is rising vertically upwards with a uniform acceleration \(a .\) If the mass of the pilot in the helicopter is \(m .\) What is the magnitude and direction of the action force cxerted by the helicopter on the surrounding air? (a) \(m(g+a)\) vertically downwards (b) \(m(g-a)\) vertically upwards (c) \((M+m)(g+a)\) vertically upwards (d) \((M+m)(g+a)\) vertically downwards

Short Answer

Expert verified
Option (d): \((M+m)(g+a)\) vertically downwards.

Step by step solution

01

Understanding the Problem

A helicopter, with mass \(M\), is rising with an acceleration \(a\). The mass of the pilot within the helicopter is \(m\). We need to find the force the helicopter exerts on the air, which requires considering both gravitational and inertial forces.
02

Identifying Forces Involved

The helicopter needs to exert a force against gravity on both its own mass \(M\) and the mass of the pilot \(m\), as well as to provide the force needed to achieve the acceleration \(a\) upwards.
03

Calculating Gravitational Force

The gravitational force acting on the combined mass of the helicopter and the pilot is \((M + m)g\), directed downwards.
04

Calculating the Required Force for Acceleration

The additional force needed to accelerate the helicopter and pilot upwards at acceleration \(a\) is \((M + m)a\).
05

Determining the Total Force by the Helicopter

The total force exerted by the helicopter must overcome gravitational force and provide the necessary upward force for acceleration: \((M+m)(g+a)\).
06

Identifying the Force's Direction

Since the helicopter is moving upwards, the force exerted by it on the air is directed upwards but as an action force according to Newton's third law, the force the helicopter exerts will be directed downwards.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Action and Reaction Forces
Newton's Third Law of Motion states that for every action, there is an equal and opposite reaction. This law is fundamental when analyzing forces like those at play with a rising helicopter.

In the case of the helicopter, it exerts a force downwards on the air (action force), and in turn, the air pushes back upwards on the helicopter (reaction force). This interaction is what allows the helicopter to ascend.

As the helicopter moves upwards, it is not just fighting against the force of gravity but also pushing air downwards, which combines to create the necessary lift force. This lift is the reaction force which is equal in magnitude and opposite in direction to the action force imposed by the helicopter. Thus, understanding action and reaction forces helps explain the mechanisms of flight and how the helicopter stays afloat.
Gravitational Force
Gravitational force is the attractive force that the Earth exerts on objects. It pulls anything with mass downwards toward the center of the Earth.

For the helicopter problem, gravitational force acts on both the helicopter and the pilot. The gravitational force can be calculated using the formula:
\[ F_{gravity} = (M + m)g \]
where:
  • \( M \) is the mass of the helicopter.
  • \( m \) is the mass of the pilot.
  • \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \) on Earth).
The effect of gravity is significant, as it acts to pull both the helicopter and pilot downwards, opposing any upward motion. This means the helicopter must exert enough force to counter this gravitational pull, in addition to any forces required for its acceleration.
Net Force Calculation
Calculating the net force involves combining the forces due to gravity and the forces required for acceleration.

In this case, to move upwards with an acceleration \( a \), the helicopter needs to exert a force to overcome the gravitational pull and to provide additional upward force.

The net force is given by:
\[ F_{net} = (M + m)(g + a) \]
Here:
  • \( (M + m)g \) is the gravitational force acting downwards.
  • \( (M + m)a \) is the force required to provide upward acceleration.
The combination of these forces, \( (M + m)(g + a) \), represents the total force the helicopter needs to exert to maintain its upward acceleration. The net force is directed upwards, providing sufficient push against the gravitational force and achieving the desired vertical motion.
Vertical Motion
Vertical motion refers to the movement of an object along the vertical axis, either upwards or downwards. In the helicopter's context, vertical motion is of primary interest because it involves counteracting gravity and providing sufficient lift.

The helicopter must apply a net upward force to rise. This requires solving for the sum of forces acting on the system, which includes gravity and the necessary lift for upward acceleration.

The vertical motion equation is given by:
\[ F_{motion} = (M + m)(g + a) \]
This equation confirms how much force is required to achieve vertical motion with a specific acceleration.
Key points about vertical motion include:
  • The necessity to counteract gravitational pull.
  • The importance of maintaining sufficient lift to achieve desired motion.
  • The relevance of Newton's laws in predicting trajectory and force requirements.
Understanding vertical motion within this framework allows us to appreciate how helicopters maneuver vertically through deliberate force management.

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Most popular questions from this chapter

An inextensible string \(A B\) is tied to a block \(B\) of negligible dimensions and passes over a small pullcy \(C\) : so that the frec end \(A\) hangs \(h_{1}\) unil above the ground on which the block \(B\) rests. In this initial position shown in figure, the frec cnd \(A\) is \(h\) unit below \(C\). If now the end \(A\) moves horizontally with a velocity \(u\), obtain an cxpression for the velocity of the block at any time \(t\). Solution In time \(t\), the end \(A\) moves to the position \(A_{1}\). So that \(A A_{1}=u l .\) The block \(B\) moves upwards to the position \(B_{1}\). Let \(B B_{1}=y\). 'Ihen length \(A_{1} C=h+y \quad\) ln the \(\Delta A C A_{1},(h+y)^{2}=h^{2}+(u i)^{2}\) \(\begin{array}{llll}\text { or } h^{2}+y^{2}+2 h y=h^{2}+u^{2} t^{2} & \therefore & y^{2}+2 h y-u^{2} t^{2}=0 & \ldots(1)\end{array}\) \Lambdafter solving equation (1), we get \(y--h+\sqrt{h^{2}+u^{2} t^{2}}\) Ihis is the equation for the displacement \(y\) of the block. Velocity of the block: \(v-\frac{d y}{d t}-\frac{d}{d t}\left[h \mid\left(h^{2} \mid u^{2} t^{2}\right)^{1 / 2}\right]-\frac{1}{2}\left(h^{2} \mid u^{2} t^{2}\right)^{-1 / 2} \times 2 u^{2} t\) or \(\quad v-\frac{u^{2} t}{\left(h^{2}+u^{2} t^{2}\right)^{1 / 2}}\)

\(\Lambda\) cat of mass \(m=1 \mathrm{~kg}\) climbs to a rope hung over a light frictionless pulley. The opposite end of the rope is tied to a weight of mass \(M=2 \mathrm{~m}\) lying on a smooth horizontal plane. What is the tension of the rope when the cat moves upwards with an acceleration \(a=2 \mathrm{~m} / \mathrm{s}^{3}\) relative to the rope'? Solution Let \(a\) be the absolute upward acceleration of the monkey and \(a\) ' be the absolute downward acceleration of the rope. \(a\) ' is also the tightward acceleration of \(M\). Then, \(b=a-\left(-a^{\prime}\right)\) (since relative acceleration is the vector difference between the absolute accelerations) or \(b-a=a^{\prime}\) Considering upward motion of the cat \(\quad T-m g=m a \ldots\) (i) Considering rightward motion of \(M\) \(T=M a^{\prime}=M(b-a) \quad \ldots(\) ii \()\) From (i) and (ii), we get \(T=\frac{m M}{m+M}(g+b)=\left(\frac{m \times 2 m}{m+2 m}\right)(10+2)=\frac{2 m}{3} \times 12=8 \mathrm{~N}\)

\(\Lambda\) trolley of mass \(100 \mathrm{~kg}\), starting from rest, describes \(100 \mathrm{~m}\) in 10 second. \Lambdat that instant, i.e., at the commencement of the \(11^{\text {th }}\) second, two packets, each of mass \(12.5 \mathrm{~kg}\), are gently placed in the trolley. I low far does it move in the next 10 seconds assuming that the forces on the trolley remain the same throughout? Solution Since \(s-u t+\frac{1}{2} a t^{2}\) \(100-0 \div \frac{1}{2} a 100 \quad \therefore \quad a-2 \mathrm{~m} / \mathrm{sec}^{2}\) The velocity at the cnd of 10 seconds \(=v=u+a t=0+2 \times 10=20 \mathrm{~m} / \mathrm{s}\) \(F=m a=100 \times 2=200 \mathrm{~N}\) During the next 10 seconds, acceleration \(-\frac{200}{125}-1.6 \mathrm{~m} / \mathrm{sec}^{2}\) The velocity al the commenecment of the 11 th socond \(v^{\prime}\) is given by \(m u=m^{\prime} v^{\prime}\) (Taw of conscrvation of momentum) \(100 \times 20=125 \times v^{\prime}\) \(\therefore \quad v^{\prime}-\frac{100 \times 20}{125}-16 \mathrm{~m} / \mathrm{scc}\) \(\therefore\) The distance travelled in the next 10 seconds \(=s^{\prime}-u t+\frac{1}{2} a t^{2}\) \(-16 \times 10+\frac{1}{2} \times 1.6 \times 100-240 \mathrm{~m}\)

A chain of mass per unit lengh \(\lambda\) and lenght \(1.5 \mathrm{~m}\) rests on a fixed smooth sphere of radius \(R-\left(\frac{2}{\pi}\right) \mathrm{m}\) such that \(A\) end of chain is at the top of sphere while the other end is hanging freely. Chain is held stationary by a horizontal thread \(P A\). The tension in the thread is (a) \(\lambda g\left(\frac{1}{2}+\frac{2}{\pi}\right)\) (b) \(\lambda g\left(\frac{\pi}{2}+\frac{2}{\pi}\right)\) (c) \(\lambda g\left(\frac{2}{\pi}\right)\) (d) None

A car starts from rest with a constant acceleration of \(2 \mathrm{~m} / \mathrm{s}^{2}\). Alter \(5 \mathrm{~s}\), a ball is dropped through the window of the ear. The window of the car is at a height of \(1.25 \mathrm{~m}\) from the ground. The acceleration of the ball after it is dropped is given by (a) \(10 \mathrm{~m} / \mathrm{s}^{2}\) (b) \(\sqrt{(10)^{2}+(2)^{2}} \mathrm{~m} / \mathrm{s}^{2}\) (c) \(12 \mathrm{~m} / \mathrm{s}^{2}\) (d) \(2 \mathrm{~m} / \mathrm{s}^{2}\)
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