# Does $F=ma$ still apply when a force creates torque?

Consider a wheel and axle. I apply a force on the edge. That creates a torque. However, why doesn't $$F=ma$$ still apply? Aren't I still applying a force to the entire wheel/axle assembly, why doesn't it accelerate linearly?

Is it because the axle supplies a reaction force such that there is no net force on the wheel axle assembly, just a torque.

My question exists because of how a cars tire works, where friction from road both creates a counter torque and a push forward. However, many times a force just creates a torque.

• Please clarify in your mind the $a$ above pertains to a point on the edge, or the center of mass? – John Alexiou Jan 6 at 14:40

If the force is directly in line with the center of mass of any unrestrained object, it will cause linear acceleration only, if the force is not directly in line with the center of mass, it will cause linear acceleration and rotation. Consider the book on the table, if you push it away from you on the center of an edge of the book (in line with it's com), it will slide across the table. If you push on it more towards a corner (not in line with it's com) it will rotate and move away.

• Why don't we recognize that force though? Like in an engine, we talk about the force and how it creates a torque thru the crankshaft. What keeps the crankshaft from moving linearly, the opposing forces from the bearings? So in that case is it just the force from the engine can only produce a torque? – Kevin C Speltz Jan 6 at 18:03
• Yes, the normal force of the bearing opposes the linear component, so you only have rotational acceleration, torque. – Adrian Howard Jan 6 at 18:20

Almost exactly right.

The axle provides a force in the opposite direction that's equal and opposite to your applied force. But it's not a reaction force to your force. It's a separate force. An analogy would be a book lying on a table. The book applies a force to the table. What keeps the table from accelerating? The normal force of the floor on the legs of the table. That force is not a reaction force to the force the book applies to the table. The reaction force of the book on the table is the normal force of the table on the book.

If the wheel were unconstrained (out is space somewhere) then $$F=ma$$ would apply as you imagine. There would be linear acceleration and angular acceleration.

• Garyp, one follow up question. Wouldn't the axle force be equal distance away from applied force so then wouldn't the torques cancel or create a couple? Why does it still create a torque around the axle or pivot point? – Kevin C Speltz Jan 6 at 14:37
• The axle is fixed in space. That means that it acts as if its mass were infinite. That puts the center of mass and center of rotation at the center of the axle. Another way to look at it: One might consider the torques around some point other than the center. This is a perfectly valid way to approach the problem. The analysis is slightly more complicated, but note that the torque due to the axle would produce zero angular acceleration. – garyp Jan 6 at 16:27
• So dumb it down for me. What you are saying is that the opposite force of the axle essentially produces no torque since it's at the center of mass and center of rotation correct? – Kevin C Speltz Jan 6 at 16:51
• The force of the axle does produce a torque around any axis other than its own, so I think a simpler way to think of it is as if the axle has infinite mass, so contributes zero to angular acceleration. This is analogous to the way we ignore the motion of the Earth when analyzing a bouncing ball. The acceleration of the Earth is essentially zero. – garyp Jan 6 at 17:52
• I guess what I am trying to know is why does friction at the bottom of a powered wheel like a car, cause both linear and angular acceleration. Where as the force of the engine on the crankshaft only causes torque...or maybe it's just the force of the engine is only able to produce a torque as the crankshaft is fixed and therefore any force from the engine to move forward is cancelled out? – Kevin C Speltz Jan 6 at 17:58

The relationship $$\boldsymbol{F} = m\, \boldsymbol{a}_{\rm CM}$$ always holds true as long the acceleration is that of the center of mass. The is true because the definition of momentum is $$\boldsymbol{p} = m\, \boldsymbol{v}_{\rm CM}$$, and force is the rate of change of momentum.

When a force is applied away from the center of mass (say point A), not only the center of mass accelerated by $$\boldsymbol{a}_{\rm CM}$$, but also the body has rotational acceleration $$\boldsymbol{\alpha}$$.

The rotational acceleration is a result of the torque $$\boldsymbol{r}_A \times \boldsymbol{F} = \mathcal{I}_{\rm CM} \boldsymbol{\alpha}$$ where $$\boldsymbol{r}_A$$ is the location of point A relative to the center of mass. Here $$\mathcal{I}_{\rm CM}$$ is the mass moment of inertia tensor at the center of mass.

If the body is initially at rest, the linear acceleration of point A is $$\boldsymbol{a}_A = \boldsymbol{a}_{\rm CM} + \boldsymbol{\alpha} \times \boldsymbol{r}$$

So now you can find the effective mass of point A by defining $$\boldsymbol{F} = m_{\rm eff}\, \boldsymbol{a}_A$$, and using the above equations

$$\boldsymbol{F} = m_{\rm eff}\, ( \boldsymbol{a}_{\rm CM} + \boldsymbol{\alpha} \times \boldsymbol{r}_A) = m_{\rm eff} \left( \tfrac{\boldsymbol{F}}{m} + \mathcal{I}_{\rm CM}^{-1} ( \boldsymbol{r}_A \times \boldsymbol{F}) \times \boldsymbol{r}_A \right)$$

In the 2D example of a tire, with $$\boldsymbol{F} = \pmatrix{F \\ 0 \\0}$$ and $$\boldsymbol{r}_A = \pmatrix{0 \\ -R \\ 0}$$ the above equation becomes

$$F = m_{\rm eff} \left( \tfrac{F}{m} + \tfrac{F R^2}{ I_{\rm CM}} \right)$$ or $$\boxed{ m_{\rm eff} = \frac{1}{ \tfrac{1}{m} + \tfrac{R^2}{I_{\rm CM}} } }$$

where $$I_{\rm CM}$$ is the component of $$\mathcal{I}_{\rm CM}$$ along the z-axis (rotation axis).

If the wheel was a uniform disk, then $$I_{\rm CM} = \tfrac{m}{2} R^2$$, and the effective mass is $$m_{\rm eff} = \tfrac{1}{3} m$$. This means the force feels only 1/3 of the total mass as the point A under the force accelerates more than the center of mass does.

• I guess I just want to know what force the axle feels. If we applied a 2newton force at 1 meter, let's say it is a wheel, would the axle technically push back with 2 newtons? – Kevin C Speltz Jan 6 at 15:32
• "push-back" implies a constraint (or that the shaft has fixed motion) as opposed to free to accelerate due to the force. So the answer is no in general. The reaction $G$ is $$G = F - m a_{\rm CM}$$ Only when the center of mass is not accelerating the forces are equal. – John Alexiou Jan 6 at 16:44
• So in my example assume we have a wheel that is balanced and who's axle is the center of mass. The. The reaction G would be equal to F, and according to garyp, the reaction being at the center produces no torque. – Kevin C Speltz Jan 6 at 16:52
• NO. Note that for a no-slip tire the acceleration of the center is $$a_{\rm CM} = \tfrac{R^2}{I_{\rm CM}} F$$ where $F$ is the force of friction. So the shaft "reaction" is $$G = F - m \tfrac{R^2}{I_{\rm CM}} F = \left(1 - \tfrac{m R^2}{I_{\rm CM}} \right) F$$ – John Alexiou Jan 6 at 16:54
• However I think above makes sense. The force on the axles in car situation would be F minus what force it took to accelerate the wheel. Atleast I think that is what you have shown? – Kevin C Speltz Jan 6 at 17:09

In the case of extended bodies $$\mathbf F = m \mathbf a$$ applies only to the centre of mass. Other points experience net torque which causes them to gain additional angular acceleration hence for such particles $$\mathbf F \neq m \mathbf a$$