9-22-08LectPHY2048

9-22-08LectPHY2048 - Homework hints: Fa fixes the...

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T T -Mg Homework hints: Free body diagram F a fixes the acceleration of m A +m B . Notice that this is also the acceleration of m A and m B individually . Now consider what force would be needed to give e.g. m B that acceleration and what provides that force. Man pulls down with F = -T , but that F is NOT a force acting on him and the cage.
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FBD θ F G mg f G Fsin θ Fcos θ The block is pulled at constant speed v with m , F and θ given. What is μ k ? Xx FF c o s fm a 0 = = fF c o s But k fN F c o s = θ yy s i n N m gm a 0 = θ+ = = Along x k F cos N μ= θ Need N Along y N mg = −θ So, k F cos μ Last class: N
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FBD θ F G mg f G Fsin θ Fcos θ Suppose instead that μ k is given but we instead ask what angle θ minimizes the force needed to pull the crate? N On first thought you might think the answer is θ = 0 since then all the force goes in the direction of motion. But notice: The retarding force of friction, f , depends on the normal force, N. In a horizontal pull ( θ = 0 ) N is equal in magnitude to the weight ( mg ) of the crate but if we take some of the weight off by adding an upward component to our pull we reduce N and thereby the frictional force retarding the motion.
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Xx FF c o s fm a 0 = = Fcos f θ = But k fN yy s i n N m gm a 0 = θ+ = = Along x Along y N mg Fsin = −θ FBD θ F G mg f G θ θ N To answer, we must solve for F in terms of θ . f F cos = θ k N F cos μ = θ k (mg Fsin ) F cos μ = θ So,
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kk Fcos mg Fsin θ=μ −μ θ Solving for F θ+μ k k F cos sin μ = θ We must now minimize the force with respect to the angle θ . To minimize we set dF/d θ = 0 and solve for θ . () 1 dF d mg cos sin dd θ + μθ θθ 2 k dF mg( 1)( sin cos ) cos sin d θ θ θ
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Equating to zero, () kk 2 k dF mg(sin cos ) d cos sin μ θ−μ θ = θ θ+μ θ 2 k mg(sin cos ) 0 cos sin μθ μ θ = θ k sin cos 0 θ= k sin cos θ=μ θ k sin cos θ = μ θ
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FBD θ F G mg f G Fsin θ Fcos θ N ( ) 1 kk tan tan θ=μ → θ= μ Hence if the optimum pulling angle is k 0 μ= k 1 (horizontal pull) ( ) 1 tan 0 0 θ == But if say the optimum pulling angle is ( ) 1o tan 1 45 θ So,
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m 1 m 2 In the problem shown in the figure (done last time for the case of no friction) there now is friction on the incline. HITT#1 (9-22-08) We do not know in which direction the system, when let go accelerates so we must guess a direction. We guess that mass m 2 drops and make accelerations in that direction positive. In that case when we write Newton’s 2 nd law for mass m 1 the force of friction comes in as, A. f k = + μ k N B. f k = μ k N + a
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A. f k = + μ k N B.
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This note was uploaded on 08/25/2011 for the course PHY 2048 taught by Professor Field during the Fall '08 term at University of Florida.

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9-22-08LectPHY2048 - Homework hints: Fa fixes the...

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