A jet transport has a weight of 2.97 x 106 N and is at rest on the runway

In summary, a jet transport has a weight of 2.97 x 106 N and is at rest on the runway. The two rear wheels are 17 m behind the front wheel, and the plane's center of gravity is 13.1 m behind the front wheel. The ground exerts a normal force of 1.31 x 106 N on the front wheel and a normal force of 3.9 x 106 N on each of the two rear wheels.
  • #1
moonlit
57
0
I have two problems for homework that I just can't seem to figure out. Can someone please help me?! Thanks.

1) A jet transport has a weight of 2.97 x 106 N and is at rest on the runway. The two rear wheels are 17 m behind the front wheel, and the plane's center of gravity is 13.1 m behind the front wheel. Determine the normal force exerted by the ground on (a) the front wheel and on (b) each of the two rear wheels.

2) A CD has a mass of 17 g and a radius of 6 cm. When inserted into a player, the CD starts from rest and accelerates to an angular velocity of 18 rad/s in 0.53 s. Assuming the CD is a uniform solid disk, determine the net torque acting on it.
 
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  • #2
1. The plane is in equilibrium. Sum the forces in the y-direction and the moments about a point and set both sums equal to zero. Hint: the 2 rear wheels see the same reaction force from the ground (you could show that explicitly by a separate moment balance in that plane (geometric plane, not airplane)).

2. Calculate the moment of inertia of the disk about its center. Determine the angular acceleration using the kinematic data given. Then use the fact that [itex] \vec{\tau}_{\rm net} = I\vec{\alpha} [/itex]
 
  • #3
For the first problem, it should be obvious, from symmetry, that the force on the two rear wheels must be the same (if it is not obvious, or you are not allowed to say that, look at the problem from side to side to get a third equation).

Let F1 be the weight (force) on the front wheel, F2 the weight on each rear wheel. The total forces must equal the weight of the entire airplane: F1+ 2F2= 2.97 x 106 (I read "2.97 x 106 N" but I assume that was supposed to be "106). The torques about the center of gravity are 1.31F1 for the front wheel and (3.9)(2F2)for the rear wheels and they must be equal. That gives you two equations in two unknowns to solve for F1 and F2.

(If you are not allowed to say that the forces on the rear wheels are "obviously" the same, set up three equations: F1+ F2+ F3= 2.97 x 106 (F2 and F3 are the forces on the two rear wheels), 1.31 F1= 3.9F2+ 3.9F3 and w F2= wF3 where w is the distance from the center line to each of the two rear wheels. "Obviously", that last equation gives F2= F3 and then you are back to the first set of equations.

In straight line motion, we have "F= ma". In rotary motion, we have "F= I a" where I is the "moment of inertia" of the disk. If you are expected to do this problem, you should know a formula for the moment of inertia of a uniform solid disk.
 
  • #4
Ok. I'm still having problems getting the second problem. I know that T=mr^2 and the net torque = the moment of inertia x angular acceleration but I still don't get it. I've tried 17 x 6^2 = 10404 but I'm not sure where to go from there for if that's even correct. Can someone show me some of the steps or something so I can understand it better? Thanks.
 
  • #5
Originally posted by moonlit
Ok. I'm still having problems getting the second problem. I know that T=mr^2 and the net torque = the moment of inertia x angular acceleration but I still don't get it. I've tried 17 x 6^2 = 10404 but I'm not sure where to go from there for if that's even correct. Can someone show me some of the steps or something so I can understand it better? Thanks.
First calculate the moment of inertia for a disk:
I = 1/2MR2

Then find the angular acceleration:
α = Δω/Δt

Then you can find the net torque that is producing the acceleration:
Τ = I α

Take care to use correct units!
 
  • #6
I must be doing something wrong. Here's what I've come up with so far:

I=1/2MR^2
1/2(17x6^2)=306

á = Äù/Ät
18/.53=33.96

Ô = I á
306 x 33.96=10391.76
 
  • #7
Originally posted by moonlit
I must be doing something wrong. Here's what I've come up with so far:

I=1/2MR^2
1/2(17x6^2)=306
You are doing fine, just be careful to use a consistent system of units. For example:
M = 0.017 kg
R = 0.06 meters

There is nothing wrong with the units you used, if you are aware of what you are doing. You used grams and centimeters; your answer for I is in gm-cm2. Unless you meant to use those units, I would advise you to stick to the usual SI units. (But do whatever your instructor does!)
 
  • #8
Ok I got the answer of 10.39x10^-4 NxM. Thanks for all your help!
 

What is the weight of a jet transport on the runway?

The weight of a jet transport on the runway is 2.97 x 106 N. This weight is the force exerted on the runway by the jet transport due to gravity.

What is the unit of measurement for the weight of a jet transport?

The unit of measurement for weight is Newtons (N). This is a unit of force in the International System of Units (SI).

What does it mean for a jet transport to be at rest on the runway?

When a jet transport is at rest on the runway, it means that it is not moving or in motion. It is stationary and its velocity is zero.

How is the weight of a jet transport on the runway calculated?

The weight of a jet transport on the runway is calculated using the formula W = mg, where W is weight, m is mass, and g is the acceleration due to gravity. The mass of the jet transport is multiplied by the acceleration due to gravity (9.8 m/s2) to calculate its weight in Newtons.

Why is the weight of a jet transport important?

The weight of a jet transport is important because it affects its performance and safety during takeoff, landing, and flight. The weight of the jet transport must be carefully managed to ensure that it does not exceed the maximum weight limit set by the manufacturer and that it is properly balanced for safe and efficient operation.

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