Problem 506 | Friction

Problem 506
A 400 lb block is resting on a rough horizontal surface for which the coefficient of friction is 0.40. Determine the force P required to cause motion to impend if applied to the block (a) horizontally or (b) downward at 30° with the horizontal. (c) What minimum force is required to start motion?

Solution 506

Infinitesimal's picture

Part (c)

Here is one alternative solution in minimizing $P=\dfrac{160}{\cos \alpha + 0.4 \sin \alpha}$.

To minimize $P$, the denominator should be maximize. So we consider $\cos \alpha + 0.4 \sin \alpha$.

1) Trigonometric identity $a\sin x + b\cos x = \sqrt{a^2 + b^2} \sin \left(\tan^{-1} \dfrac{b}{a} + x \right)$

$\cos \alpha + 0.4\sin \alpha = \sqrt{1 + 0.4^2} \sin \left(\tan^{-1} \dfrac{1}{0.4} + \alpha \right)$

But since $-1 \leq \sin x \leq 1$, $\sin \left(\tan^{-1} \dfrac{1}{0.4} + \alpha \right)$ has $1$ as the maximum value. Thus,
$\cos \alpha + 0.4 \sin \alpha = \sqrt{1 + 0.4^2}$

Hence, the minimum value of $P$ is $P=\dfrac{160}{\sqrt{1.16}}=\boxed{148.56 \, \mathrm{lb}}$

2) Cauchy-Schwarz Inequality, $(a^2 + b^2)(c^2+d^2) \geq (ac + bd)^2$

We have,
(1^2 + 0.4^2)(\cos^2 \alpha + \sin^2 \alpha) &\geq& (\cos \alpha + 0.4\sin \alpha)^2\\
(\cos \alpha + 0.4\sin \alpha)^2 &\leq& (1.16)(1)\\
-\sqrt{1.16} \leq &(\cos \alpha + 0.4 \sin \alpha)& \leq \sqrt{1.16}
Which means the maximum value of $\cos \alpha + 0.4 \sin \alpha = \sqrt{1.16}$ and it goes back to the fact that the minimum value of $P$ is $\boxed{148.56 \, \mathrm{lb}}$

Jhun Vert
Jhun Vert's picture

Thank you for sharing sir. The fastest way to solve this problem is graphical, a practical way for time constraint exams. One may consider this figure to find the Pmin.



CE student
CE student's picture

You can use Lami's Theorem also

Jhun Vert
Jhun Vert's picture

Yes, that is a straightforward solution. Thank you.

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