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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:
The magnitude of the net force is √2F.
Explanation:
Since the two particles have the same charge Q, they exert the same force on the test charge; both attractive or repulsive. So, the angle between the two forces is 90° in any case. Now, as we know the magnitude of these forces and that they form a 90° angle, we can use the Pythagorean Theorem to calculate the magnitude of the resultant net force:
Then, it means that the net force acting on the test charge has a magnitude of √2F.