Calculating Required Force for Pushing Metal Blocks | Static Friction Help

In summary: Please input the answer as a mathematical equation in the space provided.If we assume that friction is the same up an down 0.25!I got it! Thank you!Please input the answer as a mathematical equation in the space provided.
  • #1
Lily131313
12
2

Homework Statement

[/B]
In an automatic materials handeling operation, metal blocks of 1.5 kg each are pushed one at a time from bottom of a stack six blocks high. If the coefficient of friction is 0.25, determine the horizontal force required to push. There is a figure bellow.

14880326036721501401957-jpg.113720.jpg


I try to solve but my answer is different the offered solution. Don't know am i wronge or there is mistake in solution. Making me crazyy!
 

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  • #2
Best to use the homework template .

Anyway - show us what calculations you have done so far .
 
  • #3
Welcome to PF. You need to post what you've done.
 
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  • #4
Nidum said:
Best to use the homework template .

Anyway - show us what calculations you have done so far .
 

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  • #5
Sorry for color of paper!
 
  • #6
That is very hard to read . Please type out any further workings .

How many block surfaces have friction force acting on them ?

Is normal load the same for each surface ?
 
  • #7
Nidum said:
That is very hard to read . Please type out any further workings .

How many block surfaces have friction force acting on them ?

Is normal load the same for each surface ?
 

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  • #8
Lily131313 said:
Sorry for color of paper!
It's also not oriented correctly ?:)

There's not many equations on it; It would be better to type them in thus making it easier for helpers to help you (while saving them from going blind and suffering a neck injury from turning their heads sideways to see what you've written o0) :smile:)
 
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  • #9
Nidum said:
That is very hard to read . Please type out any further workings .

How many block surfaces have friction force acting on them ?

Is normal load the same for each surface ?
As you can see on figure, 6 blocks each one 1.5 kg. Bottom one is always pushed meaning for sure after that happens another one will be put on top so that again they are six.
 
  • #10
gneill said:
It's also not oriented correctly ?:)

There's not many equations on it; It would be better to type them in thus making it easier for helpers to help you (while saving them from going blind and suffering a neck injury from turning their heads sideways to see what you've written o0) :smile:)

I agree, just didn't know it flips the photo. Plus i am on phone not pc. Ill do again.
 
  • #11

@Lily131313 You may be misunderstanding! In general, we can only help you by asking questions that guide you in the right direction. So, @Nidum wasn't asking because he didn't know, but because he wanted you to think about that aspect of the problem, as he may have seen where your mistake lies.
 
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  • #12
PeroK said:
@Lily131313 You may be misunderstanding! In general, we can only help you by asking questions that guide you in the right direction. So, @Nidum wasn't asking because he didn't know, but because he wanted you to think about that aspect of the problem, as he may have seen where your mistake lies.

Ok got ur point! Hm let's try again then!
 
  • #13
What I would like to see is a diagram showing what forces you think are acting on the block . You can safely assume that it's six blocks during the pushing out phase and that a new block is only added to the stack at a later time .
 
  • #14
Nidum said:
What I would like to see is a diagram showing what forces you think are acting on the block . You can safely assume that it's six blocks during the pushing out phase and that a new block is only added to the stack at a later time .

Exactly. I thought as bottom one is always pushed all mass has to be sum as it is on top of frist one.

Fw= mg= 6*1.5*9.81=88.29 N ⬇️
 
  • #15
How many rubbing surfaces on the block being moved ?
 
  • #16
Lily131313 said:
Exactly. I thought as bottom one is always pushed all mass has to be sum as it is on top of frist one.

Fw= mg= 6*1.5*9.81=88.29 N ⬇️
 

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  • #17
Nidum said:
How many rubbing surfaces ?[/QUOTEi

Two up and down. I try that as well again didn't come to right solution.
 
  • #18
Draw a free body diagram for the moving block .
 
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  • #19
Nidum said:
Draw a free body diagram for the moving block .
If we assume that friction is the same up an down 0.25!
 

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  • #20
I got it! Thank you!
 

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FAQ: Calculating Required Force for Pushing Metal Blocks | Static Friction Help

How do I calculate the required force for pushing a metal block?

The required force for pushing a metal block can be calculated by multiplying the coefficient of static friction between the block and the surface it is on by the weight of the block. This will give you the minimum force needed to overcome static friction and start moving the block.

What is static friction?

Static friction is the force that must be overcome to start an object's motion while it is at rest. In the case of pushing a metal block, static friction is the force that must be overcome to start moving the block.

How do I determine the coefficient of static friction for a metal block?

The coefficient of static friction for a metal block can be determined experimentally by gradually increasing the force applied to the block until it starts to move. The coefficient can be calculated by dividing the force needed to overcome static friction by the weight of the block.

Are there any factors that can affect the required force for pushing a metal block?

Yes, there are various factors that can affect the required force for pushing a metal block, including the weight of the block, the coefficient of static friction, and the surface it is on. Additionally, the presence of any lubricants or contaminants on the surface can also affect the required force.

How can I decrease the required force for pushing a metal block?

To decrease the required force for pushing a metal block, you can either increase the coefficient of static friction by using a rougher surface or decrease the weight of the block. Additionally, using a lubricant or reducing any friction-inducing factors on the surface can also decrease the required force.

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