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# sol4 - HW Solutions 4 8.01 MIT Prof Kowalski Newtons third...

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HW Solutions 4 - 8.01 MIT - Prof. Kowalski Newton’s third law, forces and motion with pulleys . Note: All the figures are at the END! 1) 5.13 The natural choice of axis is x and y which is along the incline and perpendicular to the incline respectively because there is no accel- eration and two forces , −→ T and −→ N , out of 3 forces present are along these axes. Writing −→ F = 0 for both masses: a) F x acting on B is 0 T AB w sin α = 0 T AB = w sin α . b) F x acting on A is 0 T A T AB w sin α = 0 T A = w sin α + w sin α = 2 w sin α . c) F y = 0 for both A and B. N A w cos α = 0 and N B w cos α = 0 N A = N B = w cos α . 2) 5.14 a) In level ﬂight, the thrust and drag are horizontal and the lift and weight are vertical. At constant speed, the net force is zero: F x = F f = 0 (1) F y = L w = 0 (2) so F=f and w=L. b) When the plane attains the new speed, it is again in equilibrium and so the new values of the thrust and drag, F and f are related by F = f ; if F = 2 F , f = 2 f . c) The drag force f v 2 f f = v 2 v 2 v = v f f So in order to increase the the magnitude of the drag force by a factor of 2, the speed must increase by a factor of 2. 1

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3) 5.15 a) The tension is related to the masses and acceleration(it’s best to take y up for both blocks) by T m 1 g = m 1 a 1 (3) T m 2 g = m 2 a 2 (4) b) For the bricks, mass m 1 , accelerating upward, let a 1 = a 2 = a (the counterweight will accelerate down at the same rate as m 1 goes up since the string does not stretch). Subtracting the two
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