capa5 - 1) Draw free body diagrams for M2 and M1 and assign...

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1) Draw free body diagrams for M2 and M1 and assign coordinate axes: All forces in y-direction cancel out. Not so the forces in x-direction: The magnitude of the net-force F net,2 on the mass M 2 is (F-F contact ). Therefore, the mass M 2 accelerates with an acceleration ‘a’, or in Newton’s words: F net,2 = a·M 2 The net force acting on the mass M 1 is the contact force F contact . Since the two masses are in contact with each other they will speed up at the same rate (otherwise their velocity would start to differ and they would no longer be together.) Therefore, we know that the acceleration is the same for the two masses and we can write for the mass M 1 : F contact = a·M 1 Adding the two formulas together yields: F = a·(M 1 + M 2 ) From this we find the acceleration a = F / (M 1 + M 2 ) Now, we substitute ‘a’ in F contact = a·M 1 and we find F contact = F·M 1 / (M 1 + M 2 ) 2) This problem can be solved the same way as problem 1. Just exchange M 1 and M 2 . 3) Free body diagram (side view): From the free body diagram we can see that there is a non-zero net-force acting on the mass: F N 2 M 2 g F contact N 1 M 1 g F contact x y T mg x y Plane in which the rotation occurs T mg F net θ
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Since the mass is rotating in a plane we know that F net and m g form a 90 degree angle. Therefore, we can write:
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This note was uploaded on 11/07/2010 for the course PHYS 1120 taught by Professor Rogers during the Spring '08 term at Colorado.

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capa5 - 1) Draw free body diagrams for M2 and M1 and assign...

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