# Pressing a mass against a spring on a table with friction

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.

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

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

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)

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.

• 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. Commented Jul 4, 2020 at 17:43
• @gtoques I have edited my answer.i think i understand what problem you were having. Commented Jul 4, 2020 at 17:54

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.