sample1_soln_p2 - 2 6 W T = 4 The two circular disks shown...

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TAM 210/211 BL1 Name: Sample Exam1 Section: 3. If the weight of the box is known to be W (and the rings are massless), what is the tension in the rope which goes from B to D? (I am taking x positive to the right and y positive up.) We could start by breaking the forces into their components. (It takes a little extra time and is not necessary if it is not requested explicitly, but it earns us more partial credit if we make silly mistakes along the way.) j W ˆ W = v ; j T ˆ 1 1 T = v ; ( ) j i T ˆ 45 sin ˆ 45 cos 2 2 + = T v ; ( ) j i T ˆ 45 sin ˆ 45 cos 3 3 + = T v ; For an inextensible cord, the tensions are equal, therefore the magnitudes T 1 = T 4; similarly, T 2 = T 7. So, j T ˆ 1 4 T == v ; i T ˆ 2 5 T = v ; j T ˆ 6 6 T = v ; ( ) j i T ˆ 45 sin ˆ 45 cos 2 7 = T v . For ring B, applying one of the equilibrium conditions: 2 / 2 2 6 0 45 sin 2 6 0 T T T T F y = = = For ring A, 2 1 3 2 1 2 / 2 3 2 / 2 2 0 1 45 sin 3 45 sin 2 0 T T T T T T T T T F y + = + = = + = And 3 2 0 45 cos 3 45 cos 2 0 T T T T F x = = = . Using this for the above result 2 / 2 1 2 T T = , i.e. 2 / 1 6 T T = Finally, for the box, W T W T F y = = = 1 0 1 0 , so
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Unformatted text preview: 2 / 6 W T = 4. The two circular disks shown are identical. The radius of each disk is R (known) and the weight of each disk is W (also known). All surfaces are smooth, so you can neglect friction. Draw a free body diagram for each disk, and find the normal force between the top disk and the horizontal wall. Using Newton’s third law, the magnitudes N3 and N4 are equal. We don’t in the end need this to solve the problem (though if we had been asked for N1 or N2 we would have). Using the equilibrium condition for the top ball: 2 / 4 5 5 30 sin 4 N N N N F x = ⇒ = − ⇒ = ∑ ( ) 3 / 3 5 3 / 5 2 / 3 / 4 30 cos 4 W N W N W N W N F y = ⇒ = ⇒ = ⇒ = − ⇒ = ∑ 60º 2R W N3 N2 N1 30º 30º W N4 N5 30º T4 T2 W T1 FBD of the box: FBD of ring A: T3 T7 T5 FBD of ring B: T6...
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This test prep was uploaded on 04/07/2008 for the course TAM 211 taught by Professor Downing during the Fall '05 term at University of Illinois at Urbana–Champaign.

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