Chapter05

# Chapter05 - 5.1 Model We can assume that the ring is a...

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5.1. 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 F F T T T x x x x x net N ( ) = = + + = Σ 1 2 3 0 F F T T T y y y y y net N ( ) = = + + = Σ 1 2 3 0 Evaluating the components of the force vectors from the free-body diagram: T T x 1 1 = − T 2 x = 0 N T T x 3 3 30 = ° cos T 1 y = 0 N T T y 2 2 = T T y 3 3 30 = − ° sin Using Newton’s first law: + ° = T T 1 3 30 0 cos N T T 2 3 30 0 ° = sin N Rearranging: T T 1 3 30 0 8666 = ° = ( )( ) = cos . 100 N 86.7 N T T 2 3 30 0 5 = ° = ( ) ( ) = sin . 100 N 50.0 N Assess: Since r T 3 acts closer to the x -axis than to the y -axis, it makes sense that T T 1 2 > .

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5.2. 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 r F net N = 0 . This is a vector equation, so it has both x - and y -components: F T T x net N ( ) = = 3 2 0 cos θ = T T 3 2 cos θ F T T T T y net N ( ) = = = 1 3 3 1 0 sin sin θ θ We have two equations in the two unknowns T 3 and θ . Divide the y -equation by the x -equation: T T T T 3 3 1 2 1 1 60 1 60 58 0 sin cos tan . tan . . θ θ θ θ = = = = = ( ) = ° 80 N 50 N Now we can use the x -equation to find T T 3 2 = = ° = cos θ 50 N cos58.0 94.3 N The tension in the third rope is 94.3 N directed 58.0 ° below the horizontal.
5.3. 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: sin . sin . . θ θ = = = = ° 2 m 3 m 0 677 0 667 41 8 1 Newton’s first law for this situation is F F T T T T x x x x net N N ( ) = = + = ⇒ − + = Σ 1 2 1 2 0 0 cos cos θ θ F F T T w T T w y y y y y net N N ( ) = = + + = + = Σ 1 2 1 2 0 0 sin sin θ θ The x -component equation means T T 1 2 = . From the y- component equation: 2 1 T w sin θ = = = = ( ) ( ) ° = = T w mg 1 2 2 2 41 8 sin sin sin . θ θ 20 kg 9.8 m/s 196 N 1.333 147 N 2 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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5.4. 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: F F T T f x x x x x net k N ( ) = = + + = Σ 1 2 0 F F T T f y y y y y net k N ( ) = = + + = Σ 1 2 0 From the free-body diagram: T T f 1 2 1 2 1 2 0 cos cos θ θ + = k N T T 1 2 1 2 1 2 0 0 sin sin θ θ + = N N From the second of these equations T T 1 2 = . Then from the first: 2 10 1 T cos ° = 1000 N = ° = = T 1 2 10 1000 N 1000 N 1.970 508 N cos Assess: The two tensions are equal, as expected, since the two players are pulling at the same angle. The two add
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• Fall '08
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