Irodov 1.67: Angling the Pull to Save Tension – Amit & Deepali Sinha Foundation

Problem Irodov 1.67

A bar of mass $m$ sits on a plane inclined at angle $\alpha$ to the horizontal, with coefficient of friction $k$ between the bar and the incline. The bar is dragged up the slope by a thread that makes some angle $\beta$ with the incline’s own surface (not with the horizontal), lifting the thread away from the slope rather than running along it; assume the bar moves at constant velocity. Find the angle $\beta$ that minimizes the thread’s tension, and the value of that minimum tension.

The bar of mass m on the incline of angle α, dragged up the slope by a thread whose tension T makes the angle β with the incline's own surface (the dashed ray is that surface direction). — The bar of mass $m$ on the incline of angle $\alpha$ , dragged up the slope by a thread whose tension $\vect T$ makes the angle $\beta$ with the incline’s own surface (the dashed ray is that surface direction).

Given

$m$
$\alpha$
$k$

Find

\beta

that minimizes the tension

T

;

T_{\min}

Solution

Step 1Resolve forces along and normal to the incline

The same scene with the three forces this step puts on the bar: the normal N, the weight m g and kinetic friction f acting down the slope, since the bar is being dragged up it. — The same scene with the three forces this step puts on the bar: the normal $\vect N$ , the weight $m\vect g$ and kinetic friction $\vect f$ acting down the slope, since the bar is being dragged up it.

With the bar moving at constant velocity, both the along-slope and normal directions balance. Resolving the thread tension $T$ into components $T\cos\beta$ along the slope and $T\sin\beta$ away from it,

\begin{align*} T\cos\beta &= mg\sin\alpha + kN \tag{1} \\ N + T\sin\beta &= mg\cos\alpha \tag{2}\end{align*}

Step 2Eliminate the normal force

Solving the normal balance (2) for $N$ and substituting it into the along-slope balance (1) isolates $T$ in terms of $\beta$ alone.

\begin{align*}N &= mg\cos\alpha - T\sin\beta \\ T\cos\beta &= mg\sin\alpha + k(mg\cos\alpha-T\sin\beta) \\ T\cos\beta + kT\sin\beta &= mg(\sin\alpha+k\cos\alpha) \\ T(\cos\beta+k\sin\beta) &= mg(\sin\alpha+k\cos\alpha)\end{align*}

T = \dfrac{mg(\sin\alpha+k\cos\alpha)}{\cos\beta+k\sin\beta}

(3)

Step 3Minimize

T

over

\beta

The numerator of (3) does not depend on $\beta$ , so $T$ is smallest when the denominator $\cos\beta+k\sin\beta$ is largest. Writing it as a single sinusoid puts an exact bound on it.

\begin{align*}\cos\beta+k\sin\beta &= \sqrt{1+k^2}\,\cos(\beta-\varphi), \quad \tan\varphi=k \\ \left(\cos\beta+k\sin\beta\right)_{\max} &= \sqrt{1+k^2}, \text{ attained at } \beta=\varphi\end{align*}

\tan\beta = k

(4)

Step 4The minimum tension itself

Substituting the maximized denominator $\sqrt{1+k^2}$ from Step 3 into (3) gives the minimum tension directly.

T_{\min} = \dfrac{mg(\sin\alpha+k\cos\alpha)}{\sqrt{1+k^2}}

(5)

Check

At $k=0$ (a frictionless incline), Step 3 gives $\beta=0$ : pulling exactly along the slope is already optimal, because with no friction coefficient to multiply it the normal force $N=mg\cos\alpha-T\sin\beta$ costs nothing to carry and there is no benefit left to tilting the pull to lighten it, not because $N$ itself vanishes. $T_{\min}=mg\sin\alpha$ then recovers the plain along-slope result. As $k$ grows, $\beta=\arctan k$ grows too, tilting the pull further off the surface to lighten $N$ more aggressively, exactly the trade-off Step 1’s remark anticipated.

Remark

$\beta=\arctan k$ is numerically the same value as the friction angle, the incline angle at which an unsupported bar is on the verge of sliding down by itself ( $\tan\alpha=k$ , from balancing $mg\sin\alpha$ against $kmg\cos\alpha$ ). The two problems do not reduce to the same trigonometric combination, this one maximizes $\cos\beta+k\sin\beta$ , the friction angle instead solves $\sin\alpha-k\cos\alpha=0$ , so the coincidence is not a shared equation. What is shared is only the 1-to- $k$ weighting: each problem sets a term of size 1 against a term of size $k$ at some angle, and any such 1-to- $k$ combination is extremal or balanced at an angle whose tangent is $k$ , which is why $\arctan k$ keeps reappearing across Coulomb-friction problems with coefficient $k$ without the two pictures being the same physical statement (this $\beta$ is measured off the incline’s own surface, the friction angle off the horizontal).