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I recently learned that normal force often shifts from its usual line (below the COM) to counter the torque produced by friction.

Is this shifting only possible when friction is there?

Subhranil Sinha
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2 Answers2

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It happens when other torques exist and those torques are countered by the box being flat on the ground.

As an example, imagine a box on a frictionless surface. While it sits still, the normal force must act through the center of mass to counter gravity.

Now push the box sideways from one edge near the top. Because the push is above the center of mass, it creates a torque. Gravity cannot provide a restoring torque. Instead, we consider the normal force acts at a location that provides a torque sufficient to allow total torque to be zero.

BowlOfRed
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  • Why do we consider normal firce to shift? – Matt May 07 '17 at 06:47
  • If you add a horizontal force not through the COM, then it would create a torque about the COM. Although no other forces have appeared, the item does not rotate. Therefore we can consider the normal force has shifted in manner that allows the net torque to be zero. – BowlOfRed May 07 '17 at 06:52
  • Is there a physical or atomic explanation that how does normal reaction you know like kniw how much and when to shift ? – Matt May 07 '17 at 06:53
  • https://physics.stackexchange.com/questions/314729/does-the-normal-reaction-shift-when-a-force-is-applied – BowlOfRed May 07 '17 at 06:56
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The normal force is there to enforce a constraint (a block not going through the table for example). But in the most general case, there might be two normal forces in the ends of the block, or a single force and an equipollent torque.

In 2D if a force $N$ at $x=0$ acts together with a torque $\tau$ then it is said the line of action of the forces passes through $x = \frac{\tau}{F}$. So when the reaction torque is zero the line of action is "centered".

In order to find the conditions that make the reaction torque zero you need a free body diagram with all the forces acting (external, gravity, friction) and the balance equation(s).

Consider the three cases below:

ThreeBocks

All three are equivalent if

$$\begin{aligned} N & = N_1 + N_1 \\ \tau & = \ell \;\tfrac{N_1-N_2}{2}\\ x &= \frac{\tau}{ N} \end{aligned} $$

So where the normal load is said to depend on what convention from the three above is used to place normal loads in free body diagrams. Given any one pair of values from $(N_1,N_2)$, $(N,\tau)$ or $(N,x)$ the other two pairs can always be calculated using the equations above.

John Alexiou
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