The Ladder from the IPhOO

A uniform ladder of mass m m and length L \mathcal{L} is resting on a wall. A man of mass m m climbs up the ladder and is in perfect equilibrium with the ladder when he is 2 3 L \frac{2}{3}\mathcal{L} the way up the ladder. The ladder makes an angle of θ = 3 0 \theta = 30^\circ with the horizontal floor. If the coefficient of static friction between the ladder and the wall is the same as that between the ladder and the floor, which is μ \mu , what is 1000 μ 1000 \cdot \mu , expressed to the nearest integer?

678 159 567 424

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2 solutions

Maedhros 777
Dec 21, 2013

First, draw a free body diagram . We know that F 1 = μ N 1 F_1 = \mu N_1 . Since the system is at translational equilibrium, the net horizontal force is 0 and so N 2 = F 1 N_2 = F_1 . Knowing that N 2 N_2 is the normal force of the wall on the ladder, F 2 = μ N 2 = μ 2 N 1 F_2 = \mu N_2 = \mu^2 N_1 . Again, since there is translational equilibrium, the vertical forces balance out and so N 1 + F 2 = 2 m g N_1 + F_2 = 2mg : N 1 = 2 m g F 2 = 2 m g μ 2 N 1 N_1 = 2mg - F_2 = 2mg - \mu^2 N_1 μ 2 N 1 + N 1 = N 1 ( μ 2 + 1 ) = 2 m g \mu^2 N_1 + N_1 = N_1(\mu^2 + 1) = 2mg N 1 = 2 m g μ 2 + 1 N_1 = \frac{2mg}{\mu^2+1} Thus, N 2 = 2 μ m g μ 2 + 1 N_2 = \frac{2\mu mg}{\mu^2 + 1} and F 2 = 2 μ 2 m g μ 2 + 1 F_2 = \frac{2\mu^2 mg}{\mu^2+1} . Now we can use the fact that the system is at rotational equilibrium as well, so Σ τ = 0 \Sigma \tau = 0 . Using the base of the ladder as the pivot point: Σ τ = m g cos θ ( L 2 ) m g cos θ ( 2 L 3 ) + F 2 L cos θ + N 2 L sin θ = 0 \Sigma \tau = -mg\cos\theta(\frac{L}{2}) - mg\cos\theta(\frac{2L}{3}) + F_2 L\cos\theta + N_2 L\sin\theta = 0 m g cos θ 2 2 m g cos θ 3 + 2 μ 2 m g cos θ μ 2 + 1 + 2 μ m g sin θ μ 2 + 1 = 0 \frac{-mg\cos\theta}{2} - \frac{2mg\cos\theta}{3} + \frac{2\mu^2 mg\cos\theta}{\mu^2+1} + \frac{2\mu mg\sin\theta}{\mu^2+1} = 0 2 μ 2 cos θ + 2 μ sin θ μ 2 + 1 = cos θ 2 + 2 cos θ 3 = 7 cos θ 6 \frac{2\mu^2\cos\theta + 2\mu\sin\theta}{\mu^2+1} = \frac{\cos\theta}{2}+\frac{2\cos\theta}{3}=\frac{7\cos\theta}{6} 2 μ 2 cos θ + 2 μ sin θ = 7 cos θ 6 μ 2 + 7 cos θ 6 2\mu^2\cos\theta+2\mu\sin\theta = \frac{7\cos\theta}{6}\mu^2 + \frac{7\cos\theta}{6} μ 2 ( 2 cos θ 7 cos θ 6 ) + 2 μ sin θ 7 cos θ 6 = μ 2 ( 5 cos θ 6 ) + 2 μ sin θ 7 cos θ 6 = 0 \mu^2(2\cos\theta - \frac{7\cos\theta}{6}) + 2\mu\sin\theta - \frac{7\cos\theta}{6} = \mu^2(\frac{5\cos\theta}{6}) + 2\mu\sin\theta - \frac{7\cos\theta}{6} = 0 Plugging into the quadratic formula: μ = 2 sin θ ± 4 sin 2 θ 4 ( 5 cos θ 6 ) ( 7 cos θ 6 ) 7 cos θ 3 \mu = \frac{-2\sin\theta\pm\sqrt{4\sin^2\theta - 4(\frac{5\cos\theta}{6})(\frac{-7\cos\theta}{6})}}{\frac{7\cos\theta}{3}} Setting θ = 3 0 \theta = 30^\circ and discounting the negative answer (-2.06), we find that μ 0.678 \mu \approx 0.678 and so 1000 μ 1000\mu \approx * 678 *.

Excellent! I like how you correctly drew the free-body diagram, unlike others on the contest.

Ahaan Rungta - 7 years, 5 months ago

GoOd Job.

Sarah B - 7 years, 5 months ago

Solution is perfect !!!

Achint Gupta - 7 years, 5 months ago

perfect!

Vladimir Tsiolkovsky - 7 years, 2 months ago

In the translational equilibrium expression in the vertical direction, why is the force due to the weight of man equal to 2mg? Shouldn't it be mg instead?

Gen Purp - 7 years, 5 months ago

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The weight of the man is m g mg , but the ladder also weighs m g mg so N 1 + F 2 = m g + m g = 2 m g N_1 + F_2 = mg + mg = 2mg .

Maedhros 777 - 7 years, 5 months ago

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Oh thanks. i missed that the ladder too has mass m.

Gen Purp - 7 years, 5 months ago

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@Gen Purp Me too :(

Tanya Gupta - 7 years, 3 months ago
Ayon Ghosh
Jul 9, 2017

just balance the forces in the x and y components and take sum of torques 0 about the end of ladder.write everything in terms of mg. and then cancel out the m and the l from the torque equations.ladder has uniform mass density so com lies at l/2 and is conc. with mass m.we also know theta = 30.we solve for mu and get 0.678. so floor function on mu gives 678.

(p.s ahaan rungta i solved iphoo and got 6/10 [P-1,4,5,7,9,10].(including this one)what rank do i have because topper has 9/10 and rank 2 and 3 are tied at 8/10,i submitted solutions too but no reply from your side).

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