2 boxes stacked, determining forces on each other.

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In summary, the child's mother has oiled a small strip of the horizontal floor so that it is very slick; the coefficient of kinetic friction between the oiled section of floor and the lower block is only 0.12 and the coefficient of static friction is insignificantly different. Therefore, the vertical component of the contact force on the upper block by the lower block is 0.12N.
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Homework Statement


The mass of the upper block in the stack is 0.85 kg. The mass of the lower block in the stack is 1.67 kg. The coefficients of friction between the two blocks are: static 0.75, and kinetic 0.61. The child's mother, who likes to encourage his experiments, has oiled a small strip of the horizontal floor so that it is very slick; the coefficient of kinetic friction between the oiled section of floor and the lower block is only 0.12 and the coefficient of static friction is insignificantly different.

Before the pushing starts, here are some questions about the vertical forces acting on the two blocks.

b) What is the vertical component of the contact force on the upper block by the lower block? HINT: You must apply Newton's First Law.

Homework Equations



ƩF=0
Since at this point in time the boxes are stationary

The Attempt at a Solution



I know the Fg of upper box is 8.33N, I also know that along with the normal force on the upper block there is also a FB2B1 block 2 being the upper block.
I believe Fg,total= 24.696N, after many other attempts I tried to say Fg,total = NB2, therefore
ƩF = n + FB2B1 - Fg,B2
= 24.696N + FB2B1 - 8.33N = 0
this gives me a negative result for FB2B1 which I know is incorrect.
Here are my free body diagrams, I believe I may be missing something in my FBD which is why I cannot get the answer I am looking for.
 

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  • #2
Looking at the upper block, where do each of the forces come from?
 
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  • #3
I ended up getting the answer, I was over-thinking the problem as I do sometimes, especially when it's late. Thank You for your reply
 
  • #4
Well done. Would you like to provide the actual method for the benefit of others?
 
  • #5


Dear student,

Thank you for sharing your attempt at a solution and your free body diagrams. It seems like you have a good understanding of the forces at play in this scenario. However, I believe you may be overthinking the problem and making it more complicated than it needs to be.

Since the boxes are stationary, we can apply Newton's First Law which states that the net force on an object is equal to zero. This means that the forces in the vertical direction must also balance out to zero.

Let's start by considering the upper block. We know that the only forces acting on it in the vertical direction are its weight (Fg) and the normal force (n) from the lower block. Since the block is not moving, we can say that Fg = n.

Now, let's look at the lower block. It has the weight of the upper block (Fg) acting downwards, and the normal force (n) from the floor acting upwards. In addition, there is also the frictional force (Ff) acting in the opposite direction of motion. Since the floor has been oiled, the coefficient of kinetic friction is 0.12, which means that Ff = 0.12n.

Now, we can set up our equation:

ƩF = Fg + n - Ff = 0

Substituting in our values, we get:

8.33N + n - 0.12n = 0

Solving for n, we get:

n = 8.33N / 0.88 = 9.47N

Therefore, the vertical component of the contact force on the upper block by the lower block is 9.47N.

I hope this helps clarify the problem for you. Remember to always start with the simplest approach and only add in additional forces if necessary. Keep up the good work in your experiments and scientific pursuits!
 

What is the concept of "2 boxes stacked" in determining forces?

The concept of "2 boxes stacked" in determining forces refers to a scenario in which two boxes are stacked on top of each other, and the forces acting on each box are being analyzed. This can be used to understand the interaction between the two boxes and the forces that keep them in place.

How do you calculate the forces on each box in this scenario?

To calculate the forces on each box, you need to consider the weight of each box, as well as any external forces acting on them. The weight of the top box will exert a downward force on the bottom box, while the weight of the bottom box will exert an upward force on the top box. External forces, such as friction or applied forces, can also impact the forces on each box.

What is the significance of determining forces on each box in this scenario?

Determining forces on each box in this scenario can help to understand the equilibrium of the stacked boxes and ensure that the forces are balanced. It can also be used to predict how the boxes will behave under different conditions and how much weight each box can support.

Can the forces on each box change over time?

Yes, the forces on each box can change over time if there are changes in the external forces acting on them. For example, if an additional weight is placed on top of the top box, the forces on each box will change, and the bottom box may experience more compression forces.

Are there any limitations to using this method to determine forces on stacked boxes?

One limitation of using this method is that it assumes the boxes are rigid and do not deform under the applied forces. In reality, the boxes may have some flexibility, which can affect the distribution of forces. Additionally, this method does not take into account the structural integrity of the boxes, which can also impact the forces acting on them.

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