Dynamics Question #2

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  • #1
looi76
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Homework Statement


The diagram shows the forces acting on a child on a playground slide. Air resistance is negligible.
http://img353.imageshack.us/img353/6035/403jc7.png [Broken]​
[/URL]

(a) The mass of the child is 40kg.
(i) Calculate the size of the force W.
(ii) Describe the force that, along with W, makes up the 'equal and opposite pair' of forces described by Newton's third law.

(b) The size of the friction force is 90N.
(i) Calculate the component of W parallel to the slide.
(ii) Calculate the acceleration of the child.

(c) The slide is 5.5m long. If the child maintains this acceleration, calculate her speed at the bottom.

Homework Equations


[tex]F = m.a[/tex]
Frictional Force [tex]= F - mg\sin\theta[/tex]
Normal Force = [tex]N_{normal force} = mg\cos\theta[/tex]

The Attempt at a Solution


(a)(i) [tex]m = 40kg[/tex]
[tex]F = m.a[/tex]
[tex]F = 40 \times 9.81[/tex]
[tex]F = 392.4N[/tex]

(ii)?

(b)(i) Frictional Fore [tex]= 90N[/tex]
F [tex]= mg\sin\theta[/tex]
F [tex]= 40 \times 9.81 \times \sin{24}[/tex]
F [tex]= 159.60[/tex]
F [tex]= 159.60 - 90[/tex]
F [tex]= 69.6N[/tex]

Are my answers correct? I need to be sure to continue...
 
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Answers and Replies

  • #2
Hootenanny
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Homework Statement


The diagram shows the forces acting on a child on a playground slide. Air resistance is negligible.
http://img353.imageshack.us/img353/6035/403jc7.png [Broken]​
[/URL]

(a) The mass of the child is 40kg.
(i) Calculate the size of the force W.
(ii) Describe the force that, along with W, makes up the 'equal and opposite pair' of forces described by Newton's third law.

(b) The size of the friction force is 90N.
(i) Calculate the component of W parallel to the slide.
(ii) Calculate the acceleration of the child.

(c) The slide is 5.5m long. If the child maintains this acceleration, calculate her speed at the bottom.

Homework Equations


[tex]F = m.a[/tex]
Frictional Force [tex]= F - mg\sin\theta[/tex]
Normal Force = [tex]N_{normal force} = mg\cos\theta[/tex]

The Attempt at a Solution


(a)(i) [tex]m = 40kg[/tex]
[tex]F = m.a[/tex]
[tex]F = 40 \times 9.81[/tex]
[tex]F = 392.4N[/tex]

(ii)?

(b)(i) Frictional Fore [tex]= 90N[/tex]
F [tex]= mg\sin\theta[/tex]
F [tex]= 40 \times 9.81 \times \sin{24}[/tex]
F [tex]= 159.60[/tex]
F [tex]= 159.60 - 90[/tex]
F [tex]= 69.6N[/tex]

Are my answers correct? I need to be sure to continue...
Without checking your arithmetic, your methods thus far are good.
 
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  • #3
looi76
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Thnx Hootenanny,

(b)(ii) [tex]F = m.a[/tex]

[tex]a = \frac{F}{m} = \frac{69.6}{40} = 1.7475 ms^{-2}[/tex]

(c) [tex]v^2 = u^2 + 2as[/tex]
[tex]v^2 = 2 \times 1.7475 \times 5.5[/tex]
[tex]v = \sqrt{2 \times 1.7475 \times 5.5}[/tex]
[tex]v = 4.38 ms^{-1}[/tex]


Are my answers correct?
 
  • #4
Hootenanny
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Thnx Hootenanny,

(b)(ii) [tex]F = m.a[/tex]

[tex]a = \frac{F}{m} = \frac{69.6}{40} = 1.7475 ms^{-2}[/tex]

(c) [tex]v^2 = u^2 + 2as[/tex]
[tex]v^2 = 2 \times 1.7475 \times 5.5[/tex]
[tex]v = \sqrt{2 \times 1.7475 \times 5.5}[/tex]
[tex]v = 4.38 ms^{-1}[/tex]


Are my answers correct?
Again your methods are correct, but I ain't checking your arithmetic for you :wink:
 
  • #5
looi76
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Thnx Hootenanny, just wanted to make sure that in part (b)(ii) using sin was correct rather than cos?
 
  • #6
Hootenanny
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Thnx Hootenanny, just wanted to make sure that in part (b)(ii) using sin was correct rather than cos?
Yes, sine of the angle with give you the component parallel to the slope, whereas cosine of the angle will give you the component perpendicular to the slope.
 
  • #7
looi76
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The vector N in the diagram has no purpose in this question?!
 
  • #8
Hootenanny
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The vector N in the diagram has no purpose in this question?!
Nope it is not relevant for this question, it is merely shown for completeness I guess.
 

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