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icang [17]
4 years ago
9

Gravitational force on the moon is only 1/6 that of the gravitational force on earth. what would be the weight of a 10-kg object

on the moon and on earth? what would be its mass on the moon and on earth?
Physics
1 answer:
MA_775_DIABLO [31]4 years ago
6 0
According to Newton's Second Law of Motion, the force is equal to the mass of an object multiplied by its acceleration. When you talk about gravitational force, the acceleration referred to here is the acceleration due to gravity. This is very familiar to us in physics. The acceleration due to gravity on Earth is equal to 9.81 m/s². It actually depends on the location. According to the Universal Law of Gravitation:

F = Gm₁m₂/d²

The force is a factor of the product of two masses and their distance from each other. The G is a constant called the universal gravitational constants. So, gravitational force is actually a relative force exerted by one body to another. 

Going back the Second Law of Motion, we can modify the equation to:

F = mg

Since it is mentioned that the gravity on the moon is only 1/6 of the Earth, then the gravity for moon is:
g,moon = 1/6(9.81) = 1.635 m/s²
So, let's compare the weight of the object with a mass of 10 kg. The weight is actually the force due to gravity pulling you towards the center of the body.

Weight on Earth = (10 kg)(9.81 m/s²) = 98.1 N
Weight on Moon = (10 kg)(1.635 m/s²) = 16.35 N

The mass, on the other hand, is not affected by gravity. It is always constant. Therefore, the mass of the object on the moon is the same with its mass on the Earth.
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A 1300 kg steel beam is supported by two ropes. (Figure
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Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

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where R_1,R_2 are the magnitudes of the tensions in ropes 1 and 2, respectively;

  • the net vertical force acting on the beam is

R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0

where m=1300\,\rm kg and g=9.8\frac{\rm m}{\mathrm s^2}.

Eliminating R_2, we have

\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)

R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2

R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2

-R_1 \sin(50^\circ) = -\dfrac{mg}2

R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}

Solve for R_2.

\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0

\dfrac{R_2}2 = -mg\cot(110^\circ)

R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}

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the answer is A. A school crossing guard raises a stop sign that weigh 10 N

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