Applying Force On An Angle

  • Thread starter danago
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  • #1
danago
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Homework Statement


A lawnmower weighing 20kg has a handle attached at 35 degrees to the horizontal. If a man wishes to push the lawnmower, so that after 2.5s, its velcoity is 2 m/s, what force should he apply along the handle.

Homework Equations


[tex]
\sum {\overrightarrow F } = m\overrightarrow a
[/tex]

The Attempt at a Solution


ok, since after 2.5 seconds, it requires a velocity of 2 m/s, i found the acceleration of the mower to be 0.8 m/s/s. Since it has a mass of 20kg, the net force applied must then be:

[tex]
\sum {\overrightarrow F } = 20 \times 0.8 = 16N
[/tex]

So the horizontal component of the force he applies must be 16N in the direction of acceleration.

The horizontal component of the force (F) he applies down the handle will be [tex]F cos 35[/tex]. So solving for F:

[tex]Fcos35=16[/tex]

Gives F=19.5N.

According to the answer book, the answer is 120N. Where have i gone wrong?

Thanks,
Dan.
 
Last edited:

Answers and Replies

  • #2
Hootenanny
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If the question is quoted correctly, then I would agree with your answer. Does the text not mention any other factors, such as friction?
 
  • #3
danago
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Nope, no friction at all. What ive said is pretty much quoted word for word from the book.

Thanks for the reply.
 
  • #4
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yeah the book is def. wrong
 
  • #5
danago
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ok thats good to hear :) Thanks for the help
 

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