How will object's velocity change when a frictional force increases?

[Lotto]

TL;DR Summary

I apply a force $F$ on a object with a mass $m$, the object is moving with a constant velocity $v$ because a frictional force is equal to my force. How will the velocity change when the frictional force increases and my force increases on the same magnitude as well?

Let's say that the mass of the objest is suddenly bigger, so when I want to maintain the constant movement, my force must increase as well. But will the velocity have the same magnitude? I think that the velocity will be smaller, so if I want to have the same initial velocity, I must apply an additional force to accelerate it on it. Am I right? If so, how to account for that velocity decrease?

 

[Dale]

How does the mass get bigger? Mass is conserved, so it cannot just get bigger. Importantly, what is the momentum of the added mass?

 

[kuruman]

How does the mass get bigger? Mass is conserved, so it cannot just get bigger.

You can have, for example, a mass sliding along a rough table under the influence of the force and at some point drop another mass on it that sticks (magnetically) to the sliding mass.

To @Lotto : Write the equation that expresses the force $F$ needed to pull the mass at constant velocity across a surface with friction. How should you change the force when the mass suddenly increases by a factor $f$?

 

[Lotto]

You can have, for example, a mass sliding along a rough table under the influence of the force and at some point drop another mass on it that sticks (magnetically) to the sliding mass.

To @Lotto : Write the equation that expresses the force $F$ needed to pull the mass at constant velocity across a surface with friction. How should you change the force when the mass suddenly increases by a factor $f$?

The force $F$ must be equal to the frictional force, so $F=\mu mg$. So when $m'=m+f$, then my $\Delta F=\mu fg$.

 

[kuruman]

Increasing by a factor $f$ means $m\rightarrow f~m$.

 

[Lotto]

Increasing by a factor $f$ means $m\rightarrow f~m$.

OK, then $F'=fF$, so I have to increase it f-times.

 

[kuruman]

Yes.

 

[Lotto]

Yes.

But how will the velocity change? Will it be the same? Or smaller?

 

[kuruman]

That is not clear. It depends on how you model the transition. In the simplest case, you can assume that the at all times the mass is proportional to the force as they both increase. Then Newton's first law guarantees that the velocity will remain the same. If that is not the case, then if the force reaches its final value first, the final constant velocity will be greater because there will be a time interval of acceleration in the same direction as the velocity. The opposite will be the case if the mass reaches its final value first.

 

[Lotto]

That is not clear. It depends on how you model the transition. In the simplest case, you can assume that the at all times the mass is proportional to the force as they both increase. Then Newton's first law guarantees that the velocity will remain the same. If that is not the case, then if the force reaches its final value first, the final constant velocity will be greater because there will be a time interval of acceleration in the same direction as the velocity. The opposite will be the case if the mass reaches its final value first.

Let's say that I have a shovel, I put it on the ground with snow and start accumulating it on the shovel, so the mass of the snow is still getting bigger. I move with the shovel with a constant velocity. So when I want to maintain the constant velocity, do I need to apply any other force than the force that is equal to the increasing frictional force?

Don't I need to accelerate the shovel on the initial velocity because otherwise the velocity would decrease? I am not sure.

 

[russ_watters]

The way you are describing the problem, you've constrained the velocity to be constant...

 

[Lotto]

The way you are describing the problem, you've constrained the velocity to be constant...

Yes, that's the point. What do I need to do to make the velocity constant?

 

[Dale]

Let's say that I have a shovel, I put it on the ground with snow and start accumulating it on the shovel, so the mass of the snow is still getting bigger. I move with the shovel with a constant velocity. So when I want to maintain the constant velocity, do I need to apply any other force than the force that is equal to the increasing frictional force?

Don't I need to accelerate the shovel on the initial velocity because otherwise the velocity would decrease? I am not sure.

What is the horizontal velocity of the falling snow? Does it match the shovel or the ground or something else?

This is the information I asked you to provide in post #2

 

[russ_watters]

Yes, that's the point. What do I need to do to make the velocity constant?

Nothing beyond what was already described unless the model gets more complex and you remove the constraint.

 

[Lotto]

What is the horizontal velocity of the falling snow? Does it match the shovel or the ground or something else?

The snow is on the ground, it is not snowing. But beause I want to get rid of it, I use the shovel and pull it horizontally with the constant velocity. But I must increase the force since the mass of the snow on the shovel is still getting bigger. And my question is if I only must increase my force so that it is equal to the increasing frictional force or if I must besides accelerate the snow with an additional force to maintain the same velocity.

 

[Dale]

You must accelerate the snow also. The force must be equal to friction plus the rate of change of the momentum of the snow.

 

[erobz]

In the case of the snow shovel moving at constant velocity you can apply the "Momentum Equation", which states that the net force acting on the control volume is equal to the rate of momentum accumulation within the control volume plus the net outflow rate of momentum through control surface:

$$ \sum \boldsymbol{F} = \frac{d}{dt} \int_{cv} \boldsymbol{v} \rho ~ dV\llap{-} + \int_{cs} \boldsymbol{v} \rho \boldsymbol{v} \cdot d \boldsymbol{A} $$
$$ \rightarrow^+ \sum F = F_p - f_r = F_p - \mu M_{cv} g$$

The shovel ( contol volume) is moving to the right at constant speed $v$, thus in the first integral the velocity can come outside both the integral and the derivative.

$$ \frac{d}{dt} \int_{cv} \boldsymbol{v} \rho ~ dV\llap{-} = v \frac{d}{dt} \int \rho dV\llap{-} = v \dot m $$

With constant properties and uniform velocity across the control surface the last integral simplifies to:

$$ \int_{cs} \boldsymbol{v} \rho \boldsymbol{v} \cdot d \boldsymbol{A} = \cancel {\sum \dot m_o v_o}^0 - \sum \dot m_i v_i $$

There is no momentum outflow so the first term is zero and there is one inlet so you can get rid of the summation. Furthermore, the velocity of the flow crossing the control surface must be w.r.t. an inertial frame. The shovel is an inertial frame, thus:

$$ \int_{cs} \boldsymbol{v} \rho \boldsymbol{v} \cdot d \boldsymbol{A} = -\dot m (-v)$$

For the friction force the mass of the control volume is given by:

$$M_{cv} = M_{shovel} + \rho A v t $$

Putting that together I get that the force of the push $F_p$ to maintain constant velocity over time is given by:

$$F_p = 2\rho A v^2 + \mu \left( M_{shovel} + \rho A v t \right)g$$

With
$A$ as the cross section of the incoming snow
$\rho$ is the density of the snow
$\mu$ kinetic friction coefficient
$v$ velocity of the shovel (constant)