Finding Acceleration of Two Objects Touching

Question

Alex is asked to move two boxes of books in contact with each other and resting on a rough floor. He decides to move them at the same time by pushing on box A with a horizontal pushing force FP = 8.7 N. Here A has a mass mA = 10.6 kg and B has a mass mB = 7.0 kg. The contact force between the two boxes is FC. The coefficient of kinetic friction between the boxes and the floor is 0.04. (Assume FP acts in the +x direction.)

1) What is the magnitude of the acceleration of the two boxes?

I keep getting 0.47 and I can't figure out why that isn't correct. I'm a math major, this is jibberish to me, so if you could leave step by step instructions on how to solve i'd be forever grateful. Thanks for your time.

Answer 1

Acceleration = (sum of forces)/(total mass).

total mass = 17.6 kg

Push Force = 8.7 N

Frictional Force = -0.04(9.8 m/s^2)(17.6 kg)

therefore, acceleration = (8.7-0.04(9.8)(17.6))/(17.6)

Answer 2

The net force on the two boxes in contact is the vector sum of the applied ("pushing") force $ \ F_P = 8.7 \ $ N and the kinetic friction force $ \ f_k \ $ which acts in the opposite direction. The magnitude of that frictional force is 0.04 times the normal force of the force vertically upward on the boxes. Since the boxes are taken to be on a horizontal floor, the normal force equals the weight of the boxes, $ \ (m_1 + m_2)g = 17.6 \cdot 9.81 \ $ N. The kinetic frictional force is thus $ \ 0.04 \cdot 17.6 \cdot \ 9.81 = 6.9 \ $ N.

Since this is motion along one dimension, the vectors in opposition just produce a subtraction of their magnitudes, so the net force is $ \ 8.7 - 6.9 \approx 1.8 \ $ N. This is positive, so the boxes accelerate together in the positive $ \ x-$ direction, at

$$ \ a = \frac{F_{net}}{m_1+m_2} \approx \frac{1.8 \text{N}}{17.6 \text{kg.}} \approx 0.102 \frac{\text{m.}}{\text{sec.} ^2} \ . $$

On the contact forces, the lead box, B, has $ \ F_C \ $ pushing it in the positive direction and kinetic friction acting on it in the negative direction. The acceleration of the box is the common acceleration we just found. So we have

$$ F_C \ - \ 0.04 \cdot m_B \cdot g \ = \ m_B \cdot 0.102 \ \Rightarrow \ F_C \ \approx \ 2.74 + 0.71 \ \text{N} \ . $$

For the trailing box, A, the applied force $ \ F_P \ $ is in the positive direction and both the contact force and the frictional force act in the negative direction. So we have

$$ F_P \ - \ F_C \ - \ 0.04 \cdot m_A \cdot g \ = \ 8.7 \ - \ F_C - 4.16 \ = \ m_A \cdot 0.102 \ \approx \ 1.08 \ \text{N} $$

$$ \Rightarrow \ \ F_C \ \approx \ 8.7 - 4.16 - 1.08 \ \text{N} . $$

As we expect from Newton's "second law", the contact force acting on each box is equal in magnitude.