The Cable's Out - why else would there be so much STATIC?

Timed Problem Solutions: (Cannon on a spring problem)

a) Fn - mg = 0 so Fn = mg = (5200 kg)(9.8 m/s^2) = 50,960 N

b) momentum is conserved in the x-direction:
0 = m_projectilev_projectile*cos(45) - m_cannon*v

v = 3.536 m/s = final speed of cannon

c) .5*m*v^2 + 0 = 0 + .5*k*x^2
x = 1.768 m

d) F_spring = k*x = 20000 N/m * 1.768 m = 35,400 N

e) T = 2*pi*(m/k)^(1/2) = 1.99 s, so f = .5 Hz


Handout Solutions:
1.
a) I'm not drawing this diagram using ASCII characters.

b) F = 900 N

c) D = 5/9 L

2.
b) T sin(40)*(3L/4) - (800 N)*L - (600 N)*L/2 = 0

c) Px - T cos(40) = 0, so Px = 1748 N to the right ,

Tsin(40) - 800 N - 600 N - Py = 0, Py = 66.65 N directed downwards

3. From a FBD, you should be able to see that torque due to gravity opposes the torque due to the force F.

100N(0.5 m) - F*(5/3 m) = 0
F = 30 N

Book Problems:

8. Left support F1 = 8570 N, Right support F2 = 6130 N

10. Set the pivot point at a distance x from the left end.
Ma*g*x - Mc*g*(10 m - x) = 0
x = 3.0 meters

11. 3.3 meters from the adult

18. 1.1 m from pivot on side of lighter boy.

22. F1(10.0 m) – (4000 N)(8.0 m) – (3000 N)(4.0 m) – (2000 N)(1.0 m) – (250 kg)(9.80 m/s2)(5.0 m) = 0, so F1 = 5800 N

You can use the sum of forces in the vertical direction to find the other support force F2 = 5600 N

25. If we make A the support force on the left side, and B the support force for the right side, we can assume that the center of gravity is located a distance x from the right side.

Adding torques: (35.1 kg)g(170 cm – x) – (31.6 kg)g*x = 0,
x = 89.5 cm from the feet


27. T = 2500 N, Fay = 2500 N, Fax = 2500 N