Reaction force: 7 identical boxes are place next each other?

7 identical boxes are place next each other

  1. The floor is smooth

  2. Each box has a mass of 1 kg

  3. g=10

4 coefficient of friction is 0.1 (i don't know why is it given.)

5 The 7 boxes is pushed by a force of 89 N from the left side

Find the force between box 5 and box 6. ( counting from the left side)

Can you please explain how to do this, Thank you.

Note by Peter Bishop
4 years, 2 months ago

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Well, if the floor is smooth, coefficient of kinetic friction cannot be 0.1... Assuming the floor isn't smooth, let us consider the entire system of seven boxes (plus adorable massless cats) and draw its free-body diagram. Let us also assume that the force is enough to set all the boxes into motion with a constant rightward acceleration \(a\), which we need to find.

Then, we have a total leftward frictional force of \(F_{k}=\mu_{k}mg = 0.1 \times 7 kg \times 10 ms^{-2}= 7 N\). We have a total rightward applied force of \(89 N\). Since the vertical forces are balanced (because the boxes + cats are not accelerating vertically), the net force on the system is \(89 N-7 N = 82 N\) rightwards. But since we know the mass of the system, we can solve for its acceleration \(a\) using Newton's Second Law: \(F_{net} = ma \Longrightarrow a = \dfrac{F_{net}}{m} = \dfrac{82 N}{7 kg} = \dfrac{82}{7} ms^{-2}\)

So now we know that each box is accelerating rightwards with an acceleration of \(\dfrac{82}{7} ms^{-2}\). Now, let us consider Boxes 6 and 7 collectively as our system. The forces on this system (horizontal, not vertical) are the leftward frictional force (\(F_{k}=\mu_{k}mg = 0.1 \times 2 kg \times 10 ms^{-2}= 2 N\)) and the rightward reaction force \(F_{5,6}\) between Box 5 and Box 6. It is in fact this reaction force that actually accelerates Boxes 6 and 7, not the applied force.

So we can use Newton's Second Law again for this system: \(F_{net} = F_{5,6} - F_{k} = ma \Longrightarrow F_{5,6} = 2 kg \times \dfrac{82}{7} ms^{-2} + 2 N = \boxed{25.43 N}\).

Assuming that the floor is smooth, the problem becomes much simpler, due to the lack of frictional forces. The answer in that case can be easily computed in the same way.

Raj Magesh - 4 years, 2 months ago

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