9-24-08LectPHY2048

9-24-08LectPHY2048 - Homework Hint To solve you must write...

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Homework Hint To solve you must write an expression for Newton’s 2 nd law applied at the top of the hill. Do this generally, using a free-body diagram that includes a normal force, N, on the man from the seat, being careful about the signs that signify the direction of the forces and the acceleration . Now apply the condition that N = 0 at the top of the hill giving you the speed there. Now do the same at the bottom of the dip, again, being careful about the signs that signify the direction of the forces and the acceleration .
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Further problems involving friction. For the situation shown the floor has coefficient of kinetic friction μ k with the boxes of mass m 1 and m 2 ,respectively. Force F is large enough that the boxes are accelerated. Find the tension in the rope. FBD -f 1 -f 2 T -m 1 g -m 2 g -T N 1 N 2 For m 1 along y, y1 y Fm a = 11 N mg 0 = N mg = Which is used to get that, 1k 1 f μ N μ = = -Fsin θ Fcos θ
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-f 1 -f 2 T -m 1 g -m 2 g -Fsin θ -T Fcos θ N 1 N 2 For m 1 along x, x1 x Fm a = 11 x Tf m a = * For m 2 along y, y2 y a = 22 N Fsin θ mg 0 −− = N m g + Which is used to get, 2k 2 f μ N μ (Fsin θ mg ) = =+ k1 1 x T μ mg ma = For m 2 along x, x2 x a = x Fcos θ m a −= k2 2 x θ T μ θ ) ma + = *
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k1 x 1 T μ mg a m = k2 x 2 Fcos θ T μ (Fsin θ ) a m −+ = The two * equations have common accelerations a x and T . Since we are after T , we eliminate a x . Solve each for a x and equate, k 2 12 T μ θ T μ θ ) mm −− + = 2k 1 2 1 1 1 k 2 mT μ mmg mFcos θ mT m μ θ ) −=− + 1 1 2 1 1 k 1 k 2 mT mT μ θ m μ Fsin θ m μ += + 1 2 (m m )T m F(cos θμ sin θ ) + =− 1k mF(cos sin θ ) T = +
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θ m 1 m 2 μ 1 μ 2 Two boxes connected by a rigid rod are sliding down the incline. They have different coefficients of kinetic friction. Is the rod under tension or compression?
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Which of these cases holds we might anticipate depends on the relative magnitude of f 1 and f 2 . For f 1 > f 2 , box 1 holds box 2 back and we have tension (as drawn), while if f 1 < f 2 , box 2 holds box 1 back, so we have compression (force direction on each box from rod reversed meaning that the sign of each T is reversed). f 1 = μ 1 N 1 T –T -m 1 g -m 2 g θ 1 mgs in θ 2 θ θ θ f 2 = μ 2 N 2 N 2 N 1 The rod will either be applying forces to the boxes , as shown, in which case the rod is under tension, or each of those forces will be reversed in which case the rod is under compression.
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9-24-08LectPHY2048 - Homework Hint To solve you must write...

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