A box of mass 8.0kg is being dragged along the ground by a force of 30N.

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  • #1
looi76
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[SOLVED] A box of mass 8.0kg is being dragged along the ground by a force of 30N.

Homework Statement


A box of mass [tex]8.0kg[/tex] is being dragged along the ground by a force of [tex]30N[/tex].
(a) If the friction force is [tex]26N[/tex] what is the resulting acceleration?
(b) If the frictional force is one-quarter of the normal reaction force, what is the acceleration?

Homework Equations


F = m.a

The Attempt at a Solution



(a) [tex]m = 8.0kg \ , \ F = 30N[/tex]

[tex]F = 30 - 26 = 4N[/tex]

[tex]F = m.a[/tex]

[tex]a = \frac{F}{m} = \frac{4}{8} = 0.5ms^{-2}[/tex]

(b) [tex]m = 8.0kg , \ , F = 30 \times \frac{3}{4} = 22.5N[/tex]

[tex]a = \frac{F}{m} = \frac{22.5}{8.0} = 2.8ms^{-2}[/tex]

Are any answers correct?
 

Answers and Replies

  • #2
Hootenanny
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Homework Statement


A box of mass [tex]8.0kg[/tex] is being dragged along the ground by a force of [tex]30N[/tex].
(a) If the friction force is [tex]26N[/tex] what is the resulting acceleration?
(b) If the frictional force is one-quarter of the normal reaction force, what is the acceleration?

Homework Equations


F = m.a

The Attempt at a Solution



(a) [tex]m = 8.0kg \ , \ F = 30N[/tex]

[tex]F = 30 - 26 = 4N[/tex]

[tex]F = m.a[/tex]

[tex]a = \frac{F}{m} = \frac{4}{8} = 0.5ms^{-2}[/tex]
Correct :approve:
(b) [tex]m = 8.0kg , \ , F = 30 \times \frac{3}{4} = 22.5N[/tex]

[tex]a = \frac{F}{m} = \frac{22.5}{8.0} = 2.8ms^{-2}[/tex]
Careful here, notice what the question says:
If the frictional force is one-quarter of the normal reaction force, what is the acceleration?
 
  • #3
looi76
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Thnx Hootenanny,

Normal Reaction Force [tex]= 4N[/tex]

[tex]m = 8.0kg \ , \ F = 4 \times \frac{3}{4} = 3N[/tex]

[tex]F = m.a[/tex]

[tex]a = \frac{F}{m} = \frac{3}{8} = 0.4 ms^{-2}[/tex]

I don't thinks its correct, can you please explain?

By Normal reaction force, it means the reaction without the friction?
 
  • #4
Hootenanny
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By Normal reaction force, it means the reaction without the friction?
No, the http://en.wikipedia.org/wiki/Normal_force" [Broken] is the force which acts on the box and is normal to the surface. It is the force which prevents the box from entering the surface.
 
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  • #5
looi76
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I understood what normal reaction force means, but tell now I don't know how to calculate it. Can you please show me an example or explain?

Thanks...
 
  • #6
Hootenanny
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I understood what normal reaction force means, but tell now I don't know how to calculate it. Can you please show me an example or explain?
Consider your example, what are all the forcing acting in the vertical plane?
 
  • #7
looi76
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Forces acting on the vertical plane are 30N and 26N
 
  • #8
Hootenanny
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Forces acting on the vertical plane are 30N and 26N
No, they are the forces acting on the horizontal plane. What are the forces acting in the vertical plane, i.e. in the 'up and down' direction?
 
  • #9
looi76
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[tex]F = m.a[/tex]

[tex]m = 8.0kg \ , \ a = g(gravity) = 9.81ms^{-2}[/tex]

[tex]F = 8.0 \times 9.81[/tex]

[tex]F = 78.5N[/tex]

So, the normal reaction force is [tex]78.5N[/tex] ?
 
  • #10
Hootenanny
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[tex]F = m.a[/tex]

[tex]m = 8.0kg \ , \ a = g(gravity) = 9.81ms^{-2}[/tex]

[tex]F = 8.0 \times 9.81[/tex]

[tex]F = 78.5N[/tex]

So, the normal reaction force is [tex]78.5N[/tex] ?
Correct :approve:

Personally I would leave it as 78.48 N to prevent rounding errors later on.
 
  • #11
looi76
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Thnx Hootenanny, So answer (b) would be:

[tex]F = m.a[/tex]

[tex]F = 30 - \left(\frac{1}{4} \times 78.48\right) = 10.38N \ , \ m = 8.0kg[/tex]

[tex]a = \frac{F}{m} = \frac{10.38}{8.0} = 1.3ms^{-2}[/tex]

correct?
 
  • #12
Hootenanny
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Thnx Hootenanny, So answer (b) would be:

[tex]F = m.a[/tex]

[tex]F = 30 - \left(\frac{1}{4} \times 78.48\right) = 10.38N \ , \ m = 8.0kg[/tex]

[tex]a = \frac{F}{m} = \frac{10.38}{8.0} = 1.3ms^{-2}[/tex]

correct?
Correct indeed :approve:
 

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