# Is F always equal to m*a?

• B
• kent davidge
In summary, regarding Newton's second law, it is possible to consider cases where the mass changes with time, such as a rocket expelling fuel. However, in these cases, one must be careful and consider both external forces and the loss of momentum due to the mass leaving. The equation for a variable mass system is ##F_{\rm ext} + \dot m \Delta \vec v = m \vec a##, where ##\Delta\vec v## is the difference between the velocity of the exhaust and that of the object. Additionally, the rocket equation can be derived by temporarily assuming that the external force is zero and applying conservation of momentum.

#### kent davidge

Regarding Newton's second law, are we allowed to consider cases where the mass ##m## changes with time?

I see ##F = m dv/dt## so frequently that I found myself asking this question.

kent davidge said:
Regarding Newton's second law, are we allowed to consider cases where the mass ##m## changes with time?

I see ##F = m dv/dt## so frequently that I found myself asking this question.
Can you give an example of how the mass is changing with time?

Chestermiller said:
Can you give an example of how the mass is changing with time?
Perhaps a rocket expelling fuel?

The better definition of force is in terms of momentum
$$\vec F = \frac{d\vec p}{dt}.$$
For a variable mass system where ##m## is a function of ##t##, you need to be careful though. Just applying the definition
$$\vec F = \frac{d(m\vec v)}{dt} = \dot m \vec v + m \vec a$$
without considering what goes into ##\vec F## can lead you astray and the equation is not Galilei invariant (##\vec v## depends on the inertial frame). In order to get it right, you must consider both external forces that act upon the system as well as the loss of momentum due to the mass leaving. If the mass leaving does so with velocity ##\vec u##, the force equation becomes
$$\vec F_{\rm ext} + \dot m \vec u = \dot m \vec v + m \vec a,$$
where ##\vec F_{\rm ext}## is the external force acting on the system, which can be rewritten as
$$m\vec a = \vec F_{\rm ext} + \dot m (\vec u - \vec v) = \vec F_{\rm ext} + \dot m \Delta \vec v,$$
where ##\Delta\vec v## is the difference between the velocity of the exhaust and that of your object, i.e., the exhaust velocity relative to the object, which is Galilei invariant as it is the difference of two velocities.

• kent davidge and Delta2
kent davidge said:
I see ##F = m dv/dt## so frequently that I found myself asking this question.
That's the apppropriate simplification to ##F=\frac{dp}{dt}## when the mass is a constant. You see it so often because it because so many interesting and important problems involve a constant mass. The reader is expected to recognize it as the constant-mass special case without further reminders.

• kent davidge
Newton's Second Law says that "For a particle, F = m*a." By definition, a particle has constant mass and zero dimensions.

The only way I was able to derive the equation for a rocket was to first (temporarily) assume that the external force is zero, and then to apply conservation of momentum. At time t, the momentum of the rocket/fuel plus the momentum of the discharged gas is given by:
$$m(t)\mathbf{v}(t)-\int_0^t{\dot{m}(t')\mathbf{u}(t')dt'}=\mathbf{C}$$
where m(t) is the mass of rocket+fuel at time t, ##\mathbf{v}## is the velocity of the rocket+fuel, ##\dot{m}=dm/dt##, and ##\mathbf{u}##is the velocity of the discharge gas. If I differentiate this equation with respect to t, I obtain:
$$m\mathbf{a}+\dot{m}\mathbf{v}-\dot{m}\mathbf{u}=0$$
or $$m\mathbf{a}=\dot{m}(\mathbf{u}-\mathbf{v})$$

Is there another alternate derivation that is easy to make sense of.

• Delta2
As ChesterMiller said, The only way I was able to derive the equation for a rocket was to first (temporarily) assume that the external force is zero, and then to apply conservation of momentum.

That is exactly the correct way to get the rocket equation, with the exception that there is nothing temporary about the external force being zero.

Dr.D said:
As ChesterMiller said, The only way I was able to derive the equation for a rocket was to first (temporarily) assume that the external force is zero, and then to apply conservation of momentum.

That is exactly the correct way to get the rocket equation, with the exception that there is nothing temporary about the external force being zero.
I was thinking that the external force might be gravity, for example.

Dr.D said:
Newton's Second Law says that "For a particle, F = m*a."

No, it says F=dp/dt. There is no limitation to particles in the second law.

Chestermiller said:
The only way I was able to derive the equation for a rocket was to first (temporarily) assume that the external force is zero, and then to apply conservation of momentum.

And you can do that using forces because they imply conservation of momentum (due to the 3rd law).

## 1. Is F always equal to m*a?

Yes, according to Newton's Second Law of Motion, force (F) is equal to mass (m) multiplied by acceleration (a). This means that whenever an object experiences a change in acceleration, there must be a corresponding force acting on it.

## 2. Can F be equal to m*a in all situations?

In most cases, yes. However, there are certain situations where this equation may not hold true. For example, in the case of objects moving at speeds close to the speed of light, the equation may need to be adjusted to account for relativistic effects.

## 3. Can F and m*a be equal even if the object is not accelerating?

Yes, if the object is at a constant velocity, its acceleration is equal to zero. This means that the force acting on the object must also be equal to zero, as F = m*a. This is known as Newton's First Law of Motion.

## 4. How does the mass of an object affect the force and acceleration?

The greater the mass of an object, the greater the force needed to accelerate it. This is because the acceleration of an object is inversely proportional to its mass, meaning that a larger mass will require a larger force to achieve the same acceleration as a smaller mass.

## 5. Is there a limit to the force that can be exerted on an object?

Yes, there is a limit to the force that can be exerted on an object. This is known as the breaking point or yield strength of the object. If the force applied to an object exceeds its breaking point, the object will deform or break.