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Minimum force required to rotate a lamina when there is friction

2 votes

A uniform right-angled lamina is placed on a horizontal floor which is not frictionless. One of the acute angles of the lamina is $\theta$. If $F_a$ and $F_b$ are the minimum forces required to rotate the lamina about stationary vertical axes through the vertices A and B respectively, find the minimum force required to rotate the lamina about a stationary vertical axis through the vertex C.

asked Apr 6 in Physics Problems by Kartik (120 points)
edited Apr 6 by sammy gerbil
Interesting! What are your thoughts about the question?

The question is not asking about rotational acceleration. It is a question about the balance of moments : the moment provided by the applied force, and the total moment provided by friction acting uniformly over the whole interface between the lamina and floor. When these moments are balanced there is constant rotational velocity.
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1 Answer

2 votes

This is quite difficult to solve but the result is fairly simple. Possibly there is an easier solution but I do not see it.

The friction force between the lamina and the rough ground is spread uniformly over the lamina. The force per unit area is $\sigma=\frac{\mu W}{A}$ where $W, A$ are the weight and surface area of the lamina and $\mu$ is the coefficient of friction.

Using polar co-ordinates with the fixed vertex as origin, the element of area is $dA=rdrd\theta$. The force on the element is $dF=\sigma dA$. The moment about the origin of the friction force on this element is $dM=rdF$.

If $p$ is the closest distance from the origin of the side of the lamina which does not pass through the origin, then the maximum distance from the origin at polar angle $\theta$ is $p\sec\theta$. The total moment of friction about the vertex is $$M=\sigma \int \int_0^{p\sec\theta} r^2drd\theta=\frac13 \sigma p^3\int \sec^3\theta d\theta$$ $$=\frac16 \sigma p^3 [\sec\theta \tan\theta+\ln|\sec\theta+\tan\theta|]$$

See Integral of Secant Cubed. The limits of integration are $\theta_1, \theta_2$.

Suppose the triangle has vertices ABC with side lengths $a< b< c$. With the axis at vertex A we have $$p=b, \theta_1=0, \sec\theta_1=1, \tan\theta_1=0, \sec\theta_2=\frac{c}{b}, \tan\theta_2=\frac{a}{b}$$ $$M_A=\frac16 \sigma [abc+b^3\ln(\frac{a+c}{b})]$$ Likewise when vertex B is the axis $$M_B=\frac16 \sigma [abc+a^3\ln(\frac{b+c}{a})]$$ For vertex C $$p=ab/c, \sec\theta_1=a/p, \tan\theta_1=a^2/cp, \sec\theta_2=b/p, \tan\theta_2=b^2/cp$$ $$c^3M_C=\frac16 \sigma [b^3(abc+a^3\ln(\frac{b+c}{a})+a^3(abc+b^3\ln(\frac{a+c}{b}))]=a^3M_A+b^3M_B$$

The moment of friction is balanced by the moment of the applied force. When the axis is at vertices A, B the applied force is minimum when this force is applied at B, A respectively and it is perpendicular to AB; the moments are then $M_A=cF_A, M_B=cF_B$. When the axis is at vertex C the applied force is minimum when applied at A perpendicular to AC; then $M_C=bF_C$. Therefore $$bc^2F_C=a^3F_A+b^3F_B$$

In the question $\theta$ equals angle A so $$F_C=F_A\tan\theta \sin^2\theta +F_B\cos^2\theta $$

answered Apr 12 by sammy gerbil (28,806 points)
edited Apr 15 by sammy gerbil