0
$\begingroup$

Suppose I have a mass $m$ kept on a horizontal table with coefficients of friction $\mu_{s}$ (static) and $\mu_{k}$ (dynamic/kinetic). And suppose I have a spring with spring constant $k$ which is attached to a wall perpendicular to the table . Now, if I press a mass against the spring and let it go, will the mass move in the direction in which the spring moves after being released (i.e. towards its mean position)? I think it does, but I'm not able to explain this.

Now, for each displacement $x$ (after releasing the spring), the force exerted by the spring on the mass is $kx$. To get the block moving in the first place, does the force exerted on it by the spring, $kx$, need not to be greater than the max static friction $\mu_{s}mg$? But if $x$ isn't able to become sufficiently large (which because of static friction it won't), this can't happen.

And, if this is possible (which I know it is, I just can't explain it), then isn't Newton's second law violated for a little while? For very little time (till $kx > \mu_{k}mg$), my spring is accelerating the mass in the direction OPPOSITE to the kinetic friction, but $\mu_{k}mg > kx$, which means that this acceleration is in a direction opposite to the direction of net force.

I know there's a mistake in this logic, but I can't spot it.

$\endgroup$
1
  • $\begingroup$ I know that it move pretty fast. Errrm... no, you DON'T know that, unless you CALCULATE it. $\endgroup$
    – Gert
    Jul 4, 2020 at 17:39

3 Answers 3

2
$\begingroup$

Problems like these can best be approached with a Free Body Diagram (the vertical is the $y$-axis and the horizontal the $x$-axis):

Spring, mass and wall

We can see that (with no movement in the vertical direction):

$$F_N=mg$$

Since as the spring is compressed, the friction force $F_f$ points to the wall (assume the static case - index $s$):

$$F_f=\mu_sF_N=\mu_s mg$$

Now we can write using $\text{N2L}$:

$$m\mathbf{a}=\mathbf{F_S}-\mathbf{F_f}$$ Or with scalars: $$ma=-kx+\mu_s mg$$

The mass $m$ will not move for $a=0$, so:

$$\mu_s mg \geq kx$$

Or:

$$\boxed{\mu_s \geq \frac{kx}{mg}}$$

for no horizontal ($x$) motion.

But if that is not the case then (with $x$-axis motion the index $k$ is required for $\mu$):

$$ma=-kx+\mu_k mg$$

Or:

$$m\ddot{x}+kx=\mu_k mg$$

Which is the Newtonian Equation of Motion of a damped harmonic oscillator (assuming $\mu_k$ is a constant)

$\endgroup$
0
$\begingroup$

First of all if you compress a spring . It will exert a force on block towards it's mean position.If friction is present say $\mu_s m g$ , it will try to oppose the motion of block as suggested by you. In this case friction and spring force will be in opposite direction.

Now if $\mu_s m g \ge K(x)$ It will automatically decrease as static friction is variable. Therefore in this case $f_s= K(x)$ and yes block won't move.

If say block is moving : $K(x)>\mu_smg> \mu_kmg$. Thus acceleration is in direction of net force.

I think you are confusing actual length of spring to the x present in the spring force.

x is the compression/ extension of spring from the mean position.So if you even compress a spring the actual length of spring decreases but x increases.

$\endgroup$
2
  • $\begingroup$ So then the block will never be able to move right? Because for x to become large enough the block has to be able to move a distance x, which it won't be since static friction will always stop it. $\endgroup$
    – gtoques
    Jul 4, 2020 at 17:43
  • $\begingroup$ @gtoques I have edited my answer.i think i understand what problem you were having. $\endgroup$ Jul 4, 2020 at 17:54
0
$\begingroup$

first you press you the s pring x1 and $$kx1<\mu_s*m*g$$, nothing happens the two forces cancel , then you increase x so that $$kx> max(\mu_s*m*g)$$ in this moment the smaller kinetic friction sets in and you have $$kx<\mu_k*m*g$$ the force $$F=-(kx-\mu_k*m*g)$$ accelerates your mass in -x direction . I can't see where any law is broken.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.