Complex systems

32.
a) On the FBD of the bucket and girl together, the tension is directed upwards TWICE. The sum of forces in the y-direction becomes 2T - mg = 0 (for constant speed).

Solving, T = 318.5 N
b) The new tension is 350.35 N (1.1* 318.5 N). 2T - mg = ma so a = (700.7 N - 637 N)/65 kg = 0.98 m/s^2.

45.
a) Let F = 730 N and F₁₂ be the force between the two boxes.

Left block:
Fnetx = F - F₁₂ - mu*Fn1 = m₁*a
Fnety = Fn1 - m₁g = 0

Right block:
Fnetx = F₁₂ - mu*Fn2 = m₂*a
Fnety = Fn2 - m₂g = 0


Solving (2), Fn1 = m₁g and Fn2 = m₂g

Substituting:
F - F₁₂ - mu*m₁g = m₁*a
F₁₂ - mu*m₂g = m₂*a

Solving this system, a = 2.476 m/s^2, and F₁₂ = 434 N

61.
a)
We basically derived this in class for the 2nd problem. Let positive be defined to be for m₂ dropping.

a = (m₂g - m₁g*sin(theta))/(m₁ + m₂)

b)
If the system is to accelerate in the positive direction as defined above, then the net force in the numerator of the fraction above must be positive. This will occur if m₂g > m₁g*sin(theta). The system will accelerate to the right if the inequality goes the other way.

69.
This is similar to problem 46 from the other day, as well as to the do-now problem on Thursday. Here, you know mu, delta_x, and are asked to find v0. Check out your work for Problem 46 for guidelines on solving this.

81.
a) The key to this problem is knowing that the block WAS in equilibrium until the last bit of sand is added - from this, you know that Fs = Fsmax = mu*Fn. The hanging bucket (and sand) is in equilibrium, so you can find that T = m₂g.

You should obtain that the mass of the hanging bucket m₂ = 12.6 kg. You must subtract the mass of the bucket to get the sand mass = 11.6 kg.

b)
The equations in the x-direction are now equal to m₁*ax and m₂*ax respectively.

Solving the system of equations, ax = .879 m/s^2.

Problem on page:
acceleration of all three objects is 1.63 m/s^2 in the direction that moves the bucket upwards. The tension is 57.2 N.