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Published byBrycen Markes Modified about 1 year ago

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r = 600 mm C A B 200 g O Problem A small 200-g collar C can slide on a semicircular rod which is made to rotate about the vertical AB at the constant rate of 6 rad/s. Determine the minimum required value of the coefficient of static friction between the collar and the rod if the collar is not to slide when (a) = 90 o, (b) = 75 o, (c) = 45 o. Indicate in each case the direction of the impending motion.

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Solving Problems on Your Own Problem r = 600 mm C A B 200 g O A small 200-g collar C can slide on a semicircular rod which is made to rotate about the vertical AB at the constant rate of 6 rad/s. Determine the minimum required value of the coefficient of static friction between the collar and the rod if the collar is not to slide when (a) = 90 o, (b) = 75 o, (c) = 45 o. Indicate in each case the direction of the impending motion. 1. Kinematics: Determine the acceleration of the particle. 2. Kinetics: Draw a free body diagram showing the applied forces and an equivalent force diagram showing the vector ma or its components.

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Solving Problems on Your Own Problem r = 600 mm C A B 200 g O A small 200-g collar C can slide on a semicircular rod which is made to rotate about the vertical AB at the constant rate of 6 rad/s. Determine the minimum required value of the coefficient of static friction between the collar and the rod if the collar is not to slide when (a) = 90 o, (b) = 75 o, (c) = 45 o. Indicate in each case the direction of the impending motion. 3. Apply Newton’s second law: The relationship between the forces acting on the particle, its mass and acceleration is given by F = m a. The vectors F and a can be expressed in terms of either their rectangular components or their tangential and normal components. Absolute acceleration (measured with respect to a newtonian frame of reference) should be used.

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Problem Solution r = 600 mm C A B 200 g O Kinematics. r = 600 mm C A B O anan r sin a n = (r sin ) 2 a n = (0.6 m) sin ( 6 rad/s ) 2 a n = 21.6 sin m/s 2

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Kinetics; draw a free body diagram. Problem Solution r = 600 mm C A B 200 g O (0.2 kg)(9.81 m/s 2 ) O N F ma n = (0.2) 21.6 sin = 4.32 sin N =

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Problem Solution Apply Newton’s second law. (0.2 kg)(9.81 m/s 2 ) O N F ma n = (0.2) 21.6 sin = 4.32 sin N = + F t = 0: F (9.81) sin = sin cos F = 0.2 (9.81) sin sin cos + F n = ma n : N (9.81) cos = 4.32 sin sin N = 0.2 (9.81) cos sin 2 F = N For a given , the values of F, N, and can be determined

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Problem Solution (0.2 kg)(9.81 m/s 2 ) O N F ma n = (0.2) 21.6 sin = 4.32 sin N = Solution: (a) = 90 o, F = N, N = 4.32 N, = (down) (b) = 75 o, F = N, N = 4.54 N, = (down) (c) = 45 o, F = N, N = 3.55 N, = (up)

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