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On a test I used $F=m*a$ to calculate the weight of an object on Earth, using $9.8$ $m/s^2$ for $a$. I got the right answer, but the question was marked as wrong because I used the wrong formula and I should have used $W=m*g$.

Aren't the two formulas the same, only that one is "specialized" to calculate weight? What am I missing?

The question was similar to "If an item has mass x, what is its weight on Earth?"

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    $\begingroup$ Someone was picky to you. Just tell him: I decided to use F instead of W and a instead of g. Why? Because I prefer to use F for Weight and a for gravitational acceleration near the surface of earth, that’s why. $\endgroup$ – J. Manuel Mar 9 '18 at 9:11
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You are interpreting $F = m * a$ incorrectly. First of all, it should be $\Sigma F = m * a$. The equation says that all forces applied to a mass $m$ will cause the mass to accelerate by $a$. The left hand side should include all relevant forces per the specific problem. The right hand side is "fixed" as $m*a$.

In your case $W=m*g$ is a force, caused by gravitation. Not to be confused with $m*a$.

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Your weight is equal to the force exerted on you by gravity, which (assuming that you're near the surface of the Earth) you can calculate with $W=mg$, where $g=9.8 \ m/s^2$. This tells you nothing about your actual acceleration $a$, because your weight is the same whether

  1. You are sitting still, in which case $a=0$, or
  2. You are in an upward-accelerating elevator, in which case $a>0$, or
  3. You are falling off of your roof, in which case $a<0$

(where I have defined $a>0$ to be upward).

Assuming it has not already been prescribed, to calculate your acceleration you need more information - you need to know the other forces acting on the object. It's a totally different question from calculating the object's weight, despite the apparent similarity in the formulae.

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I make the case that there is insufficient difference between $F = ma$, and $W = mg$ to have deprived you of marks in that test.

Firstly, imagine a rocket with a flat front, firing away, but pinned against a sideways wall. Or imagine one of the recent electric drones, that has flown into a ceiling and is stuck there. In both cases there will be no motion. Acceleration is $a = F/m$, where F is due to thrust.

In classical mechanics, how is this fundamentally different from acceleration due to gravity? For all its differences from a rocket engine or a drone propeller, the earth is simply thrusting us towards its centre, is it not?

Secondly, there is the moon happily pulling upwards a part of the earth's ocean that it happens to be above, accelerating it by gravity but also causing motion. The moon generally varies the gravitational pull on earth. In terms of Newtonian mechanics, nothing sacred, just another force.

Thirdly, we have the kilogram-force: "The kilogram-force...is is equal to the magnitude of the force exerted by one kilogram of mass in a 9.80665 $m/s^{2}$ gravitational field" (Ref). Kilogram-force is no different from weight in magnitude. However, it can act in any direction. According to the ref, while newtons $N$ is standard, the kilogram-force is still used in China and by the European Space Agency.

$W = mg$ is merely a special case of $F = ma,\quad$.

There is no conceptual difference between the two.

I grant you permission to use my explanation to confront your teacher(s).

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I think, it is a question of being very precise and following conventional notation. What was the exact context?

If the question was: "What is the weight of an object of mass $m$ on Earth?", and you answered: "The object's weight $F$ is $F =m \cdot a$ where $a = 9.81 m/s²$" you would be correct.

I think being precise is an important aspect in physics but this looks exaggerated to me...

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    $\begingroup$ $F = ma$ is conventionally used for net force and net acceleration. Gravitational force has an equivalent acceleration on the surface; but using $F = ma$ implies that the object actually accelerates at $g$. $\endgroup$ – JMac Mar 8 '18 at 15:41
  • $\begingroup$ As I said, this seems to be about convention. To my eyes, it all depends on the scope of the question but I would not dismiss the answer as absolutely wrong. $\endgroup$ – lmr Mar 8 '18 at 17:21
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They do seem similar and I disagree that you should have lost points, but perhaps when using $F = ma$, it's implied that you're calculating a net force from a net acceleration or vice versa. When calculating weight maybe it's understood better by your teacher when you're using the formula. It does come from newtons law of universal gravitation after all ($F=K*(m_1*m_2)/r^2$ where you're only interested in the attraction between two objects and not other forces.

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