# Which acts first, friction or tension?

Suppose a block of mass $m$ and coefficient of static friction with floor $\mu$, is kept on the surface and tied to a massless inextensible string whose other end is tied to an immovable wall. (Edit: The string is already taut.) A force $F$ less than $\mu mg$ is applied on the block in the opposite direction of string? Will the tension in string be zero? If $F$ is greater than $\mu mg$, is the frictional force by the floor $\mu mg$?

What I actually mean to ask is, does friction act before the tension by the string? If so, why?

Will the answer change if instead of the wall there is a block of mass $m_1$ and coefficient of static friction $\mu _1$?

A similar case can be thought of the block kept on the floor and against the wall or another block. Will the scenario with normal force from the wall be the same as it was with tension in the previous case?

Edit: All books consider the friction to act first. I want to know the reason behind this.

It depends on how the string is set up, of course.

If there is no substantial tension in the string, then we say it is "slack" as it sags down slightly. It will only become not-slack as it stretches taut and then subtly lengthens; it will then experience some tension as a function of how long it starts and how long it gets. To a crude approximation this force is $$T(x) = \frac{Y~A}{L_0}~(x - L_0) \text{ if } L_0 < x < L_\text{max} \text{ else } 0.$$ It is zero for $x < L_0$ because the string only supports tension, not compression; it is zero for $L_\text{max} < x$ because the string breaks. In practice as the string gets closer to breaking it usually gets stiffer as it stretches out, which would be something like $k_1~(x - L_0) + k_2 (x - L_0)^2 + \dots$ but if we are nowhere near the breaking point then we do not need this extra mathematics. Technically we can of course combine together $k = Y A/L_0$ into a "spring constant" but I have written it this way because $Y$ is a material parameter called the "Young's modulus" if $A$ is the cross-sectional area of the string.

So if we start with some $x < L_0$ as you are apparently trying to do, this tension is going to be zero and the only force will come from the static friction. Once your force $F$ exceeds $\mu_\text{s}~F_\text{N}$ then the block will slide along the surface; we call this "slipping" motion. This will continue until $T(x) = F$ which will happen at $x = L_0 + \frac{L_0~F}{Y~A},$ then static friction will also be in play.

If you manage to lower $F$ while the block is moving, you could even potentially get $F < \mu_\text{s}~F_\text{N}$ while the block is moving, since kinetic coefficients of friction are usually much lower than static coefficients of friction. In this case you can remove the force and the static friction and tension would oppose each other holding the string taut; if you leave $F$ alone then the tension would be stronger than the static friction threshold and it would pull the block back the opposite direction.

• What if the string is already taut? – Shubhraneel Pal Aug 13 '17 at 2:46
• If the string starts already with some tension in it, then it's being opposed by static friction. Your force initially reduces the static friction to 0 (when your force matches the pre-existing tension) and then increases the static friction in the opposite direction until $F= T + \mu_\text s~F_\text N.$ Only at that point does static friction cease and the string can elongate, changing the tension in the string. – CR Drost Aug 13 '17 at 6:05

According to the horizontal equilibrium there are mainly 3 forces acting on the block. They are,