Irodov 1.271: Where the Floor Really Pushes Back – Amit & Deepali Sinha Foundation

Problem Irodov 1.271

A uniform cube of edge $a$ sits on a horizontal plane; the coefficient of friction between cube and plane is $k$ . The cube is launched along the plane with some horizontal velocity, slides without tipping or turning, and is brought to rest by friction. Take the fixed axis that lies in the plane and runs horizontally at right angles to the direction of the cube’s motion. The cube’s angular momentum about that axis is nonzero while it moves and zero when it stops: account for that disappearance. Then find the horizontal distance $\Delta x$ between the vertical line of action of the resultant of the gravitational forces on the cube and the vertical line of action of the resultant of the supporting plane’s normal reaction.

The cube sliding right with momentum m v at the center of mass, a height a/2 above the reference axis through O, which lies in the plane and runs into the page (the cross). — The cube sliding right with momentum $m\vect v$ at the center of mass, a height $a/2$ above the reference axis through $O$ , which lies in the plane and runs into the page (the cross).

Given

Find

an account of how the angular momentum about the stated axis disappears, and the offset

\Delta x

Solution

Step 1The angular momentum that has to go somewhere

Let $m$ be the cube’s mass and $v$ its speed. The cube translates without rotating, so every part of it carries the same velocity and the whole body’s angular momentum about the reference axis is the moment of its total momentum $mv$ , applied at the center of mass. The figure gives that moment arm as the height of the center of mass above the plane.

L = mv\cdot\frac{a}{2}

(1)

This is nonzero while the cube moves and zero once it stops, so a torque about that same axis must have acted throughout.

Step 2How fast it disappears

The vertical balance fixes the total normal reaction, and Newton’s second law along the motion then gives the deceleration. Differentiating (1) turns that into the rate at which $L$ is lost.

\begin{align*}N &= mg \\ m\frac{dv}{dt} &= -kN = -kmg \\ \frac{dv}{dt} &= -kg\end{align*}

\frac{dL}{dt} = \frac{ma}{2}\frac{dv}{dt}

\frac{dL}{dt} = -\frac{1}{2}kmga

(2)

The forces on the sliding cube: the friction f along the plane opposing v, the weight m g at the center of mass, and the resultant normal reaction N, displaced a distance Δ x from the vertical through the center. — The forces on the sliding cube: the friction $\vect f$ along the plane opposing $\vect v$ , the weight $m\vect g$ at the center of mass, and the resultant normal reaction $\vect N$ , displaced a distance $\Delta x$ from the vertical through the center.

Step 3Which torque can supply it

Take moments about the reference axis, which lies in the plane. Friction acts along the plane, so its line of action meets the axis and its moment vanishes identically; the weight and the normal reaction are both vertical and both have moment arms equal to their horizontal distances from the axis. With $x_G$ the horizontal coordinate of the center of mass and $x_N$ that of the normal resultant, the total moment is a difference of two nearly canceling terms.

\frac{dL}{dt} = mg\,x_G - N x_N

The vertical balance has already fixed $N = mg$ , so the two vertical forces carry the same coefficient and only the difference of their positions survives.

\frac{dL}{dt} = mg\,(x_G - x_N)

(3)

Friction has dropped out, so the loss of angular momentum can only be paid for by the weight and the normal reaction failing to be collinear.

The couple of the weight and the offset normal removes it.

Step 4The size of the offset

Setting the moment (3) equal to the rate (2) and canceling the common factor $mg$ leaves the displacement alone.

\begin{align*}mg\,(x_G-x_N) &= -\frac{1}{2}kmga \\ x_N - x_G &= \frac{1}{2}ka\end{align*}

\Delta x = \frac{1}{2}ka

(4)

The normal’s resultant sits ahead of the center, toward the motion.

Check

The result is a coefficient times a length, so it is a length, and it is independent of the speed and the mass, as it must be for a deceleration that is itself independent of both. At $k=0$ it gives $\Delta x=0$ : with no friction there is no deceleration, the angular momentum never changes, and the normal reaction’s resultant sits squarely under the center. At $k=1$ it gives $\Delta x=a/2$ , putting the resultant exactly at the leading bottom edge, which is the furthest forward the base can push at all. That is the same threshold as the tipping condition, and it says the analysis is self-consistent: a coefficient any larger would demand a normal resultant outside the base, and the cube would tumble forward instead of sliding.

Remark

The plane’s total reaction is the normal $\vect N$ and the friction $\vect f$ together, and because the cube does not rotate that combined resultant must have zero moment about the center of mass, so its line of action passes exactly through the center. That line is tilted from the vertical by the friction angle, whose tangent is $k$ , so between the base and the center of mass, a height $a/2$ , it drifts horizontally by $ka/2$ : the answer again, read straight off a triangle. The same picture explains the physical mechanism, which is a redistribution of pressure rather than any new force: the base presses harder near its leading edge, and it is that asymmetry, not friction itself, that carries the angular momentum away.