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Chapter 6

# Chapter 6 - 6.1 Model Visualize We can assume that the ring...

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6.1. Model: We can assume that the ring is a single massless particle in static equilibrium. Visualize: Solve: Written in component form, Newton’s first law is ( ) net 1 2 3 0 N x x x x x F F T T T = Σ = + + = ( ) net 1 2 3 0 N y y y y y F F T T T = Σ = + + = Evaluating the components of the force vectors from the free-body diagram: 1 1 2 0 N x x T T T = − = 3 3 cos30 x T T = ° 1 0 N y T = 2 2 y T T = 3 3 sin30 y T T = − ° Using Newton’s first law: 1 3 cos30 0 N T T + ° = 2 3 sin30 0 N T T ° = Rearranging: ( )( ) 1 3 cos30 100 N 0.8666 86.7 N T T = ° = = ( )( ) 2 3 sin30 100 N 0.5 50.0 N T T = ° = = Assess: Since 3 T G acts closer to the x -axis than to the y -axis, it makes sense that 1 2 . T T >

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6.2. Model: We can assume that the ring is a particle. Visualize: This is a static equilibrium problem. We will ignore the weight of the ring, because it is “very light,” so the only three forces are the tension forces shown in the free-body diagram. Note that the diagram defines the angle . θ Solve: Because the ring is in equilibrium it must obey net 0 N. F = G This is a vector equation, so it has both x - and y -components: ( ) net 3 2 cos 0 N x F T T θ = = 3 2 cos T T θ = ( ) net 1 3 3 1 sin 0 N sin y F T T T T θ θ = = = We have two equations in the two unknowns 3 T and . θ Divide the y -equation by the x -equation: ( ) 1 3 1 3 2 sin 80 N tan 1.6 tan 1.6 58 cos 50 N T T T T θ θ θ θ = = = = = = ° Now we can use the x -equation to find 2 3 50 N 94 N cos cos58 T T θ = = = ° The tension in the third rope is 94 N directed 58 ° below the horizontal.
6.3. Model: We assume the speaker is a particle in static equilibrium under the influence of three forces: gravity and the tensions in the two cables. Visualize: Solve: From the lengths of the cables and the distance below the ceiling we can calculate θ as follows: 1 2 m sin 0.677 sin 0.667 41.8 3 m θ θ = = = = ° Newton’s first law for this situation is ( ) net 1 2 1 2 0 N cos cos 0 N x x x x F F T T T T θ θ = Σ = + = ⇒ − + = ( ) net 1 2 1 2 0 N sin sin 0 N y y y y y F F T T w T T w θ θ = Σ = + + = + = The x -component equation means 1 2 . T T = From the y- component equation: 1 2 sin T w θ = ( ) ( ) 2 1 20 kg 9.8 m/s 196 N 147 N 2sin 2sin 2sin41.8 1.333 w mg T θ θ = = = = = ° Assess: It’s to be expected that the two tensions are equal, since the speaker is suspended symmetrically from the two cables. That the two tensions add to considerably more than the weight of the speaker reflects the relatively large angle of suspension.

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6.4. Model: We can assume that the coach and his sled are a particle being towed at a constant velocity by the two ropes, with friction providing the force that resists the pullers. Visualize: Solve: Since the sled is not accelerating, it is in dynamic equilibrium and Newton’s first law applies: ( ) net 1 2 k 0 N x x x x x F F T T f = Σ = + + = ( ) net 1 2 k 0 N y y y y y F F T T f = Σ = + + = From the free-body diagram: 1 2 k 1 1 cos cos 0 N 2 2 T T f θ θ + = 1 2 1 1 sin sin 0 N 0 N 2 2 T T θ θ + = From the second of these equations 1 2 . T T = Then from the first: 1 2 cos10 1000 N T ° = 1 1000 N 1000 N 508 N 2cos10 1.970 T = = = °
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Chapter 6 - 6.1 Model Visualize We can assume that the ring...

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