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How to assess whether the slope is more than 45 degrees?

  • Thread starter scarecrow1
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  • #1

Homework Statement


A 80kg skier has a force of 200 Newtons exerted on him down a slope. Assess whether the slope is less than or greater than 45 degrees.

Homework Equations


Weight = mass times acceleration due to gravity

The Attempt at a Solution



The vertical component of his weight is 800 Newton’s approximately. The video showing the problem is actually here in number 2. However the man’s solution doesn’t include all necessary trigonometry I feel.
 

Answers and Replies

  • #2
SammyS
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Homework Statement


A 80kg skier has a force of 200 Newtons exerted on him down a slope. Assess whether the slope is less than or greater than 45 degrees.

Homework Equations


Weight = mass times acceleration due to gravity

The Attempt at a Solution



The vertical component of his weight is 800 Newton’s approximately. The video showing the problem is actually here in number 2. However the man’s solution doesn’t include all necessary trigonometry I feel.
I would say that the triangle is drawn incorrectly, and then, as you say, the trigonometry is wrong.

What needed to be asked was, "What component of the weight is parallel to the slope?"
 
  • #3
I didn’t say that the trigonometry is wrong just that the explanation is incomplete. The cosine of the angle (45 degrees) should somehow be weight divided by square root of 2.
 
  • #4
SammyS
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I didn’t say that the trigonometry is wrong just that the explanation is incomplete. The cosine of the angle (45 degrees) should somehow be weight divided by square root of 2.
Sorry if I misinterpreted what you said in the OP.

A triangle needs to be drawn so that its hypotenuse corresponds to the weight, mg .
 
  • #5
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The Attempt at a Solution



The vertical component of his weight is 800 Newton’s approximately. The video showing the problem is actually here ... in number 2. However the man’s solution doesn’t include all necessary trigonometry I feel.
From what I can see in your snapshot, the triangle shown on the video is the standard 'special angle' triangle for a 45 degree angle from which we determine sin(45) = cos(45) = 1/√2.

The slope will be less than 45 degrees if the skier's given weight component parallel to the slope is less than 800 sin(45). What they seem to be showing on the video is 800 sin(45) > 200 which is the same thing.

It may be a bit misleading to refer to the "vertical component" of weight. Weight is a vertically acting force. The relevant components are those parallel and perpendicular to the slope given by mg sin(θ) and mg cos(θ) respectively. Note that the angle formed between weight vector and slope is the complement of the slope angle.
 
  • #6
Thanks Neil Parker. You certainly went beyond what was explained in the video.
 

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