FINAL_S

FINAL_S - Ph1a: Solution to the Final Exam Alejandro...

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Ph1a: Solution to the Final Exam Alejandro Jenkins, Fall 2004 Problem 1 (10 points) - The Delivery A crate of mass M , which contains an expensive piece of scientific equipment, is being delivered to Caltech. The delivery truck has a freight bed of length L (see the figure), with a coefficient of static friction μ s and a coefficient of kinetic friction μ k . Rather than move the heavy crate himself, the driver tilts the truck bed by an angle θ (see figure) and then drives the truck forward with increasing acceleration a , until the crate begins to slide. M θ x y a L g µ , µ s k For this problem, use the x, y coordinate system shown in the figure. These coordinates are fixed with respect to the truck’s bed, not to the ground. (a) (2 points) Draw a free-body force diagram for the crate in the truck’s frame of reference. Solution: See Figure 1. N is the normal force that the truck’s bed exerts on the crate, and f is the corresponding friction. θ x y f Mg N Ma Figure 1: Free-body diagram for the crate in the truck’s frame of reference
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(b) (3 points) Write down Newton’s second law for the motion of the crate in the x - and y - directions, just before it begins to slip. Solution: While the crate remains on the truck’s bed, a y = 0. Thus: Ma y = N - Mg cos θ + Ma sin θ = 0 . The maximum static friction is given by f max = μ s N . This must be the magni- tude of the friction just before the crate begins to slip. Therefore: Ma x = f max - Mg sin θ - Ma cos θ = μ s N - Mg sin θ - Ma cos θ = 0 . (c) (2 points) Determine the minimum acceleration a min for which the crate will begin to slip. Express your answer in terms of the constants shown in the figure. Solution: If we solve for a in the equations obtained in part (b) for the crate just before it begins to slip, the result must correspond to a min . Therefore a min = g μ s cos θ - sin θ μ s sin θ + cos θ . When the truck reaches a min and the driver notices the crate beginning to slide, he continues at that constant acceleration. (d) (3 points) Find the speed of the crate along the x - direction when the crate leaves the truck bed. Neglect the size of the crate. You may leave your answer in terms of a min . Solution: If the crate is sliding, then the friction is kinetic rather than static. By Newton’s second law, the acceleration is: a x = μ k ( g cos θ - a min sin θ ) - g sin θ - a min cos θ . Since this acceleration is constant, we can find the final velocity of the crate by using the equation v 2 f = 2 | a x | L . Notice that, in our coordinate system, a x for the sliding crate is negative, so that | a x | = - a x . Thus we have that v f = q 2 L [ g sin θ + a min cos θ - μ k ( g cos θ - a min sin θ )] .
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FINAL_S - Ph1a: Solution to the Final Exam Alejandro...

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