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The problem is a solid horizontal friction movement. We have an initial force $F_0$ that pushes the solid only to create the movement, at time $ t < 0$ $s$. At the time $t = 0$ $s$, the movement starts with an initial speed $v_0$ due to the initial force.

Here is a little sketch: enter image description here

Where $g$ is the acceleration of gravity, $P$ the weight, $R$ the support reaction, $f_s$ the static force friction, $\mu_s$ the static friction coefficient, $f_d$ the dynamic force friction, and $\mu_d$ the dynamic friction coefficient.

In order to solve the time equation of the movement and know when and where the solid stops, I have to know the initial velocity of the movement at $t = 0$ $s$.

Indeed, after using the second Newton's law $\sum F = m* a$ ($m$ the mass is constant), I arrived to $v_x(t) = \frac{- \mu_d R}{m}t + C$ and $x(t) = \frac{- \mu_d R}{2m}t² + Ct + D$, where $C$ is in fact the initial speed $C = v_0$, and $D = 0$ because we suppose that $x(t=0) = 0$.

Do I need to know the exact time of application of the initial force on the solid to know the initial velocity? That is to say if $F_0$ is applied during $t_{force} = 1$ $s$ on the solid I have: $v_0 = \frac{F_0*t_{force}}{m}$. Could someone tell me if I am right or explain to me what I forgot or did wrong?

Thanks in advance.

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    $\begingroup$ "We have an initial force $F_0$ that pushes the solid only to create the movement, at time $ t < 0$ $s$." This statement is a bit vague. Is $F_0$ supposed to equal the maximum possible static friction force where motion is impending? When motion begins, is the force maintained at the value $F_0$ (the maximum static friction force? $\endgroup$
    – Bob D
    Feb 6 at 17:18
  • $\begingroup$ The movement is initiated by the $F_0$. When the movement starts $t = 0$ $s$. Yes $F_0$ is enough to exceed the static friction force because $f_s = 0.6*5*9.81 = 29.43 N < F_0 = 100 N$. When motion begins, there is no force $F_0$ anymore. There is only the dynamic friction force left. $\endgroup$
    – Jejouze
    Feb 6 at 18:54
  • $\begingroup$ OK, then I can post an answer $\endgroup$
    – Bob D
    Feb 6 at 18:57

2 Answers 2

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Do I need to know the exact time of application of the initial force on the solid to know the initial velocity?

Yes, you need to know the time $t$ of application of the force $F_0$.

That is to say if $F_0$ is applied during $t_{force} = 1$ $s$ on the solid I have: $v_0 = \frac{F_0*t_{force}}{m}$. Could someone tell me if I am right or explain to me what I forgot or did wrong?

Your expression for the velocity $v_x(t) = \frac{- \mu_d R}{m}t + C$ is correct, but I'm not sure how you got the value for the initial velocity $v_o$. I'm not sure, but I think you might not have used the net force on the body which is $F_{0}-\mu_{d}mg$ for the acceleration of the body to determine the initial velocity after a specified time of application. To explain:

To get the body moving, $F_0$ has to equal the maximum possible static friction force, or

$$F_{0}=\mu_{s}mg$$

Once the body starts to move, the friction becomes kinetic or dynamic friction, which is generally considered less than static. See the friction plot below.

Now, assuming $F_0$ remains at the maximum static friction force, the body will have an acceleration of

$$a=\frac{F_{net}}{m}=\frac{\mu_{s}mg-\mu_{d}mg}{m}=(\mu_{s}-\mu_{d})g$$

Then the initial velocity after the applied force is removed and deceleration begins, where $t_{o}$ is the time the force is applied, will be

$$v_{o}=at_{o}=(\mu_{s}-\mu_{d})gt_{o}\tag{1}$$

Now the equation of motion of the body after removal of the applied force will be

$$v(t)=v_{o}-\frac{\mu_{d}mgt}{m}=v_{o}-\mu_{d}gt$$

Substituting for $v_{o}$

$$v(t)=(\mu_{s}-\mu_{d})gt_{o}-\mu_{d}gt$$

The time for the body to stop ($v(t)=0$) is then

$$t=\frac{t_{o}(\mu_{s}-\mu_d)}{\mu_d}$$

Then, in my opinion, to get the motion, $F_0$ needs to be greater than the maximal static friction force: $F_{0}\gt f_s$, no?

No. $F_{0}$ equal to the maximum possible static friction force is the minimum applied force needed to start the body moving. As I said above the precise force where friction changes from static to kinetic is not defined so you need to use the maximum possible static friction force to do the calculation. You can, of course, apply anything greater if you wish, but it is not necessary to get the body moving and will only result in a greater acceleration and greater value of $v_o$ for a given application time.

Finally, in my opinion, because the solid is at the beginning at rest, we have to take the static friction force Into consideration and not the kinetic one, so $v_o$ should be like

$$v_{o}=\frac{(F_{0}-\mu_{s}mg)t}{m}$$

What do you think?

No, because once the body starts moving static friction is no longer relevant because friction has transitioned from static to kinetic which is lower as shown in the friction plot below. So your second term in the parenthesis should be $\mu_{d}mg$.

If when the transition occurs the applied force is maintained at the level when the body started moving, then $F_{0}=\mu_{s}mg$. With these changes your equation will be

$$v_{o}=\frac{(\mu_{s}mg-\mu_{d}mg)t}{m}=(\mu_{s}-\mu_{d})gt$$

Which is the same as my equation (1) where I set $t=t_o$ for the time the force is applied, so as not to confuse it with variable $t$ in the equation of motion after the applied force is removed and the body decelerates.

Hope this helps.

enter image description here

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  • $\begingroup$ Thank you for your answer. Indeed, I forgot to consider the maximum friction force to evaluate $v_0$, thanks to remind me that! Then, in my opinion, to get the motion, $F_0$ needs to be greater than the maximal static friction force: $F_0 > f_s$, no? If it is equal, both forces compensate each other. Finally, in my opinion, because the solid is at the beginning at rest, we have to take the static friction force Into consideration and not the kinetic one, so $v_0$ should be like $v_0 = \frac{(F_0 - \mu_s mg) t_{force}}{m}$. What do you think? $\endgroup$
    – Jejouze
    Feb 7 at 11:13
  • $\begingroup$ No, you have it incorrect. I will be back to explain, but at the moment have things I have to do. $\endgroup$
    – Bob D
    Feb 7 at 11:44
  • $\begingroup$ @Jejouze I have updated my answer to respond to you. $\endgroup$
    – Bob D
    Feb 7 at 13:38
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Yes, you do need to know the exact time of application of the initial force ($F_0$) on the solid to know the initial velocity ($v_0$). Greater the time the force is applied for, greater is the change in velocity. Your derivation for the initial velocity of the object ($v_0=\frac{F_0\cdot t_{force}}{m}$) is correct if you assume that there is no kinetic friction acting. In this case, if $F_0>\mu_sN$, then the initial velocity can be given as $$v_0=\frac{(F_0-\mu_k N)\cdot t_{force}}{m}$$ where $\mu_s$, $\mu_k$ and $N$ is the coefficient of static friction, coefficient of kinetic friction and the normal reaction respectively.

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