A classical mechanics problem by Vishal Gupta

Classical Mechanics Level 2

On a winter day in Maine, a warehouse worker is shoving boxes up a rough plank inclined at an angle α \alpha above the horizontal. The plank is partially covered with ice, with more ice near the bottom of the plank than near the top, so that the coefficient of friction increases with the distance x x along the plank: μ = A x \mu = Ax , where A A is a positive constant and the bottom of the plank is at x = 0 x=0 . (For this plank the coefficients of kinetic and static friction are equal: μ k = μ s = μ \mu _k = \mu_ s = \mu .) The worker shoves a box up the plank so that it leaves the bottom of the plank moving at speed v 0 v_0 . Show that when the box first comes to rest, it will remain at rest if __________ \text{\_\_\_\_\_\_\_\_\_\_} .

v o 2 ≥ 3 g sin ⁡ 2 α A cos ⁡ α {{v}_{o}}^{2} \ge \frac {3g \sin^{2} {\alpha}} {A\cos{\alpha}} v o 2 ≤ 3 g sin ⁡ 2 α A cos ⁡ α {{v}_{o}}^{2} \le \frac {3g \sin^{2} {\alpha}} {A\cos{\alpha}} v o 2 ≠ 3 g sin ⁡ 2 α A cos ⁡ α {{v}_{o}}^{2} \ne \frac {3g \sin^{2} {\alpha}} {A\cos{\alpha}}

Vishal Gupta by Vishal Gupta , 21, India

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1 solution

So, firstly, the object will come to rest only when it's weight will be balanced by the frictional force applied on it. Let this happen at a distance x o {x}_{o} from the base. By balancing forces, we can say that for an object of mass M .

μ M g cos ⁡ α ≥ M g sin ⁡ α \mu Mg\cos{\alpha} \ge Mg\sin{\alpha}

Substituting the value of μ = A x o \mu = A{x}_{o} , we can get:

x o ≥ sin ⁡ α A cos ⁡ α {x}_{o} \ge \frac {\sin{\alpha}} {A\cos{\alpha}}

Now, at any distance x x , we can say that the retardation provided to the body is μ g cos ⁡ α + g sin ⁡ α \mu g \cos{\alpha}+ g\sin{\alpha} . Or:

a = A x g cos ⁡ α + g sin ⁡ α a = Axg\cos{\alpha} + g\sin{\alpha}

which implies:

v d v d x = − ( A x g cos ⁡ α + g sin ⁡ α ) v \frac {dv}{dx} =- (Axg\cos{\alpha} + g\sin{\alpha})

Multiplying d x dx on both the sides and integrating within proper limits, we get:

v o 2 = A g x o 2 cos ⁡ α + 2 g x o sin ⁡ α {{v}_{o}}^{2} = Ag {{x}_{o}}^{2} \cos{\alpha} + 2g{x}_{o}\sin{\alpha}

Now, using the inequality we wrote earlier, we can easily see that:

A g x o 2 cos ⁡ α ≥ g sin ⁡ 2 α A cos ⁡ α Ag {{x}_{o}}^{2} \cos{\alpha} \ge \frac {g \sin^{2} {\alpha}} {A\cos{\alpha}}

and 2 g x o sin ⁡ α ≥ 2 g s i n 2 α A cos ⁡ α 2g{x}_{o}\sin{\alpha} \ge \frac {2gsin^{2} {\alpha}} {A\cos{\alpha}}

Adding the last two equations and writing that as v o 2 {{v}_{o}}^{2} , we get:

v o 2 ≥ 3 g sin ⁡ 2 α A cos ⁡ α {{v}_{o}}^{2} \ge \frac {3g \sin^{2} {\alpha}} {A\cos{\alpha}}

P.S, I think the options should be changed to v o 2 {{v}_{o}}^{2} , instead of v o {v}_{o} . :P

5 Helpful 0 Interesting 0 Brilliant 0 Confused

a good solution my friend

Vishal Gupta - 5 years, 1 month ago

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