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

This preview shows pages 1–3. Sign up to view the full content.

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 =0and 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 ± ;i f F ± =2 F , f ± =2 f . c) The drag force f v 2 f ± f = v ± 2 v 2 v ± = v ² 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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
This is the end of the preview. Sign up to access the rest of the document.

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online