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coldgirl [10]
2 years ago
11

A baggage handler at an airport applies a constant horizontal force with magnitude F1 to push a box, of mass m, across a rough h

orizontal surface with a very small constant acceleration a.
The baggage handler now pushes a second box, identical to the first, so that it accelerates at a rate of 2a.
How does the magnitude of the force F2 that the handler applies to this box compare to the magnitude of the force F1 applied to the first box?
Physics
1 answer:
Vikentia [17]2 years ago
6 0

Answer:

F_2 = 2F_1 - F where F is the friction force that is the same for both boxes

Explanation:

So the same boxes at the same mass m would create a same friction force, let this force be F. The net force in the first and 2nd cases would be

- 1st case: F1 - F

- 2nd case F2 - F

According to Newton 2nd law, this net would make acceleration

- 1st case a_1 = (F_1 - F)/m

- 2nd casea_2 = (F_2 - F)/m

Since a_2 = 2a_1

(F_2 - F)/m = 2(F_1 - F)/m

F_2 - F = 2F_1 - 2F

F_2 = 2F_1 - F

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A uniform beam XY is 100 cm long and weighs 4.0N.The beam rests on a pivot 60 cm from end X. A load of 8.0 N hangs from the beam
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<h2>The magnitude of force F is 18N</h2>

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Moment = Force * perpendicular distance

Clockwise moments;

The force that acts clockwise is the unknown Force F and 4N force. If the  beam rests on a pivot 60 cm from end X and a Force F acts on the beam 80 cm from end X, the perpendicular distance of the force F from the pivot is 80-60 = 20cm and the perpendicular distance of the 4N force from the pivot is 60-50 = 10cm

Moment of force F about the pivot = F * 20

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Sum of clockwise moment = 40+20F...(1)

Anticlockwise moment;

The  8N will act anticlockwisely about the pivot.

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