1.
2 answers:
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Answer:
Explanation:
Given that,
Weight of jet
W = 2.25 × 10^6 N
It is at rest on the run way.
Two rear wheels are 16m behind the front wheel
Center of gravity of plane 10.6m behind the front wheel
A. Normal force entered on the ground by front wheel.
Taking moment about the the about the real wheel.
Check attachment for better understanding
So,
Clock wise moment = anti-clockwise moment
W × 5.4 = N × 16
2.25 × 10^6 × 5.4 = 16•N
N = 2.25 × 10^6 × 5.4 / 16
N = 7.594 × 10^5 N
B. Normal force on each of the rear two wheels.
Using the second principle of equilibrium body.
Let the rear wheel normal be Nr and note, the are two real wheels, then, there will be two normal forces
ΣFy = 0
Nr + Nr + N — W = 0
2•Nr = W—N
2•Nr = 2.25 × 10^6 — 7.594 × 10^5
2•Nr = 1.491 × 10^6
Nr = 1.491 × 10^6 / 2
Nr = 7.453 × 10^5 N
Answer:
A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.
Explanation:
At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:
(1)
Where:
- Period of rotation of the satellite, measured in seconds.
- Distance of the satellite with respect to the center of the planet, measured in meters.
- Gravitational constant, measured in newton-square meters per square kilogram.
- Mass of the Earth, measured in kilograms.
Now we clear the distance of the satellite with respect to the center of the planet:
(2)
If we know that ,
and
, then the distance of the satellite is:
The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:
(3)
Where:
- Mass of the satellite, measured in kilograms.
- Force exerted on the satellite by the Earth, measured in newtons.
If we know that ,
,
and
, then the gravitational force is:
A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.