Irodov 1.66: Fixed Base, Free Angle – Amit & Deepali Sinha Foundation

Problem Irodov 1.66

A small body $A$ is released from rest at the top of a wedge whose horizontal base has fixed length $l = 2.10$ m; the body slides down the wedge’s inclined face, whose angle $\alpha$ with that base can be chosen freely, and the sliding contact between the body and the incline carries a friction coefficient of $k = 0.140$ . Find the angle $\alpha$ that makes the sliding time to the bottom the smallest, and find that minimum time.

Body A released from rest at the top of a wedge of fixed base l; the incline angle α is free, and v marks the direction of sliding. — Body $A$ released from rest at the top of a wedge of fixed base $l$ ; the incline angle $\alpha$ is free, and $v$ marks the direction of sliding.

Given

$l = 2.10\text{ m}$
$k = 0.140$

Find

The angle

\alpha

that minimizes the sliding time, and that minimum time

Solution

Step 1Slope length from the fixed base

The base $l$ of the wedge is held fixed while $\alpha$ is free, so it is the base, not the height, that stays constant as $\alpha$ varies.

The slope the body travels has length

L(\alpha) = \frac{l}{\cos\alpha} \tag{1}

Step 2Acceleration down the incline

The free body diagram below resolves the three forces on $A$ against the slope.

The forces on body A: the normal N, the weight m g, and kinetic friction f acting up-slope. — The forces on body $A$ : the normal $\vect N$ , the weight $m\vect g$ , and kinetic friction $\vect f$ acting up-slope.

Resolving normal to the surface gives the normal force, and the kinetic friction the contact supplies is $k$ times it, directed up the slope since the body slides down.

N = mg\cos\alpha, \qquad f = kN = kmg\cos\alpha

Newton’s second law along the slope, taking down-slope as positive, is

mw = mg\sin\alpha - kmg\cos\alpha.

Dividing through by the mass leaves the along-slope acceleration, which the mass has dropped out of.

w(\alpha) = g(\sin\alpha - k\cos\alpha) \tag{2}

Step 3Time to reach the base

The body starts from rest under a constant acceleration, so the distance it covers in time $t$ is $wt^2/2$ , and the sliding time is fixed by setting that equal to the slope length.

L(\alpha) = \frac{1}{2}w(\alpha)t^2

Solving for $t^2$ leaves it as the ratio of the two quantities the previous steps found.

t^2(\alpha) = \frac{2L(\alpha)}{w(\alpha)}

Substituting the slope length (1) and the acceleration (2) in turn gives the sliding time as a function of the angle alone.

t^2(\alpha) = \frac{2l}{g\cos\alpha(\sin\alpha - k\cos\alpha)}

(3)

Step 4Minimizing the time

$t$ is smallest exactly where $t^2$ is smallest, so it is enough to maximize the denominator’s variable part,

f(\alpha) = \sin\alpha\cos\alpha - k\cos^2\alpha = \frac{1}{2}\sin2\alpha - \frac{k}{2}(1+\cos2\alpha).

Differentiating with respect to $\alpha$ and setting the result to zero,

\frac{df}{d\alpha} = \cos2\alpha + k\sin2\alpha = 0,

gives the condition on the optimal angle

\tan2\alpha = -\frac{1}{k}

(4)

Since $k>0$ the right side is negative, and the physical root has $2\alpha$ in the second quadrant (the one where $f$ is a maximum rather than a minimum), so

\alpha = 90^\circ - \frac{1}{2}\arctan\frac{1}{k} = 90^\circ - \frac{1}{2}\arctan\frac{1}{0.140} = 90^\circ - \frac{1}{2}(82.0^\circ) \approx 49^\circ.

Step 5The minimum time itself

At the angle satisfying (4), $\sin2\alpha = 1/\sqrt{1+k^2}$ and $\cos2\alpha = -k/\sqrt{1+k^2}$ (the unique pair in the second quadrant with that tangent), so

f(\alpha) = \frac{1}{2}\bigl[\sin2\alpha - k - k\cos2\alpha\bigr] = \frac{1}{2}\left[\frac{1+k^2}{\sqrt{1+k^2}} - k\right] = \frac{1}{2}\Bigl[\sqrt{1+k^2} - k\Bigr].

Substituting into (3) with $\cos\alpha(\sin\alpha-k\cos\alpha) = f(\alpha)$ gives the minimum time in closed form,

t_{\min} = 2\sqrt{\frac{l}{g\left(\sqrt{1+k^2}-k\right)}}.

Evaluating with $l = 2.10$ m, $k = 0.140$ , $g = 9.8\text{ m/s}^2$ : $\sqrt{1+k^2} = 1.0098$ , so $\sqrt{1+k^2}-k = 0.870$ , and

\alpha \approx 49^\circ, \qquad t_{\min} = 2\sqrt{\dfrac{2.10}{9.8\times0.870}}\text{ s} \approx 1.0\text{ s}

(5)

Check

In the frictionless limit $k\to0$ , (4) gives $\tan2\alpha\to-\infty$ , so $2\alpha\to90^\circ$ and $\alpha\to45^\circ$ : with no friction the fastest fixed-base wedge is the familiar $45^\circ$ one, the classical result recovered as the special case $k=0$ .

Remark

Because $\tan2\alpha=-1/k$ always lands $2\alpha$ in the second quadrant, the optimal angle sits between $45^\circ$ and $90^\circ$ for every $k>0$ : friction can only push the fastest wedge steeper than the frictionless case, never shallower. That is the trade-off the whole problem turns on: flattening the wedge lengthens the slope $L=l/\cos\alpha$ that must be covered, while steepening it grows the along-slope acceleration $g(\sin\alpha-k\cos\alpha)$ , and friction, by eating into that acceleration, tips the balance toward the steeper choice every time.