131_PartUniversity Physics Solution

# 131_PartUniversity Physics Solution - Applying Newtons Laws...

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Applying Newton°s Laws 5-3 5.6. I DENTIFY : Apply Newton°s 1st law to the car. The forces are the same as in Example 5.5. S ET U P : The free-body diagram is sketched in Figure 5.6. E XECUTE : x x F ma = cos sin 0 T n α α = cos sin T n α α = y y F ma = cos sin 0 n T w α α + = cos sin n T w α α + = Figure 5.6 The first equation gives cos . sin n T α α = Use this in the second equation to eliminate n : cos cos sin sin T T w α α α α + = Multiply this equation by sin : α 2 2 (cos sin ) sin T w α α α + = sin T w α = (since 2 2 cos sin 1 α α + = ). Then cos cos sin cos . sin sin n T w w α α α α α α = = = E VALUATE : These results are the same as obtained in Example 5.5. The choice of coordinate axes is up to us. Some choices may make the calculation easier, but the results are the same for any choice of axes. 5.7. I DENTIFY : Apply m = F a ! ! to the car. S ET U P : Use coordinates with x + parallel to the surface of the street. E XECUTE : 0 x F = gives sin T w α = . 2 3 sin (1390 kg)(9.80 m/s )sin17.5 4.10 10 N F mg θ = = ° = × . E VALUATE : The force required is less than the weight of the car by the factor sin α . 5.8. I DENTIFY : Apply Newton°s 1st law to the wrecking ball. Each cable exerts a force on the ball, directed along the cable. S ET U P : The force diagram for the wrecking ball is sketched in Figure 5.8. Figure 5.8 E XECUTE : (a) y y F ma = cos40 0 B T mg ° − = 2 4 (4090 kg)(9.80 m/s ) 5.23 10 N cos40 cos40 B mg T = = = × ° ° (b) x x F ma = sin40 0 B A T T ° − = 4 sin40 3.36 10 N A B T T = ° = × E VALUATE : If the angle 40 ° is replaces by 0 ° (cable B is vertical), then B T mg = and 0. A T =

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5-4 Chapter 5 5.9. I DENTIFY : Apply m = F a ! ! to the object and to the knot where the cords are joined. S ET U P : Let y + be upward and x + be to the right.
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