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nata0808 [166]
2 years ago
7

Two forces are acting on a 2.0 kg object that moves with acceleration 6.0 m/s2 in the positive y-direction. If one of the forces

acts in the positive x-direction and has magnitude of 22 N, what is the magnitude of the other force (in N)
Physics
1 answer:
Strike441 [17]2 years ago
3 0

Answer:

12N

Explanation:

We are given that one of the forces are acting only in the horizontal x-direction. As a force must be applied on an object of mass in order to cause acceleration, the 6.0ms^-2 acceleration is due to the non-horizontal force acting on the 2.0kg object.

Using Newton's Second Law of motion; we know that for a constant mass, force is equal to mass times acceleration, F=ma.

Assuming the other force is acting only in the vertical direction (question doesn't specify, thus we are finding the minimum force to cause this acceleration):

F= 2.0kg * 6.0ms^-2

F=12.0 kgms^-2

F=12 N

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Answer:

Explanation:

Given that,

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Taking moment about the the about the real wheel.

Check attachment for better understanding

So,

Clock wise moment = anti-clockwise moment

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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

6 0
3 years ago
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