What Angle Is Needed to Simulate 0.8g Deceleration on an Inclined Plane?

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In summary: Here is a little explanation about how to use it:When you see a post you want to quote, simply click the "Reply" button, which will open a reply form with the post quoted in it. Then you can write your response under the quote.If you want to quote multiple posts in one reply, click the "Insert Quotes" button at the bottom of the reply form. You can then select the posts you want to quote and click "Quote these posts" at the bottom of the page. This will insert all the selected posts into your reply form for you to respond to.In summary, the conversation discusses a truck driver's problem with calculating the angle required to resist a deceleration force of 0.8
  • #1
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A Physics Problem 02.jpg


I am a truck driver, the law requires me to restrain my load of 100kg, to resist a forward braking force (deceleration) of 0.8g.

If, I took the 100kg load and tilted it (inclined plane), what angle of tilt, would be required to represent a force of 0.8g ?

How does one calculate the angle required to equate to a force of 0.8g, I happen to know the answer is 53 degrees, but, I need to know how this is calculated.

I am a real truck driver with a real problem and look forward to your solution.
 
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  • #2
Do you know any trigonometry ?
 
  • #3
Lint said:
View attachment 79449

I am a truck driver, the law requires me to restrain my load of 100kg, to resist a forward braking force (deceleration) of 0.8g.

If, I took the 100kg load and tilted it (inclined plane), what angle of tilt, would be required to represent a force of 0.8g ?

How does one calculate the angle required to equate to a force of 0.8g, I happen to know the answer is 53 degrees, but, I need to know how this is calculated.

I am a real truck driver with a real problem and look forward to your solution.

Units are important here. After looking at the values on the diagram, I would say that if you want to work with SI units, then the force is really ##80g## Newtons, not ##0.8g## Newtons.
Now that everything is in SI units:
If you're on an inclined plane, your weight could be split into two components (using trigonometry): one parallel to the plane and the other perpendicular to the plane. We only care about whatever happens parallel to the plane for now. Given this information, do you now understand how the angle is obtained?
 
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  • #4
MohammedRady97 said:
Units are important here. After looking at the values on the diagram, I would say that if you want to work with SI units, then the force is really ##80g##, not ##0.8g##.
Quibble: If units are important then the force is not 80g either. The force of a 0.8 g acceleration on a 100 kg box is 80 kilogram-g. But this is trigonometry and what really matters are dimensonless ratios like 0.8. The units all cancel anyway.
 
  • #5
I take it we have now moved into just spoon-feeding answers instead of helping folks figure out how to get answers for themselves.
 
  • #6
phinds said:
I take it we have now moved into just spoon-feeding answers instead of helping folks figure out how to get answers for themselves.

I just edited my answer.
 
  • #7
phinds said:
Do you know any trigonometry ?

Yes, I have a little knowledge of trig, I am aware that c^2 = q^2 = b^2 and that Sine Theta = Opposite / Hypotenuse and that Cos Theta = Adjacent / Hypotenuse.
 
  • #8
l hope that was a typo and you meant c^2= a^2+ b^2 .
 
  • #9
Indeed, I did Sir, my thumbs are too big !
 
  • #10
MohammedRady97 said:
Units are important here. After looking at the values on the diagram, I would say that if you want to work with SI units, then the force is really 80g 80g Newtons, not 0.8g 0.8g Newtons.
Now that everything is in SI units:
If you're on an inclined plane, your weight could be split into two components (using trigonometry): one parallel to the plane and the other perpendicular to the plane. We only care about whatever happens parallel to the plane for now. Given this information, do you now understand how the angle is obtained?

Yes, this is my working -
 

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  • #11
phinds said:
I take it we have now moved into just spoon-feeding answers instead of helping folks figure out how to get answers for themselves.

This is nothing to do with spoon feeding anything, it's to do with providing the most eloquent answer to a specific question, any help you are able to offer is very much appreciated.
 
  • #12
Lint said:
This is nothing to do with spoon feeding anything, it's to do with providing the most eloquent answer to a specific question, any help you are able to offer is very much appreciated.
You are missing the point of this forum, which is not here for the purpose of "providing the most eloquent answer to a specific question" but rather for the purpose of helping people figure out how to GET answers to their questions. We're not giving away fish, we're teaching people how to fish.
 
  • #13
You are missing the point of this forum, which is not here for the purpose of "providing the most eloquent answer to a specific question" but rather for the purpose of helping people figure out how to GET answers to their questions. We're not giving away fish, we're teaching people how to fish.

I came to this forum for assistance to solve my problem, my knowledge of maths and trig is limited, as I am sure your knowledge of the trucking industry is limited, but if you were ever to ask me which fuel filter should be used for a Detroit 600 diesel engine, I would quite happily tell you.

If you are able to show me how to solve my problem then, I would very much appreciate your help.
 
  • #14
Lint said:
I came to this forum for assistance to solve my problem, my knowledge of maths and trig is limited, as I am sure your knowledge of the trucking industry is limited, but if you were ever to ask me which fuel filter should be used for a Detroit 600 diesel engine, I would quite happily tell you.

If you are able to show me how to solve my problem then, I would very much appreciate your help.

If I were looking for an answer about trucking, and I just wanted the answer with not expectation of ever having to solve my problem again, then I'd seek out a forum which exists for the purpose of answering such questions, not one that exists for the purpose giving explanations of how trucks work and how to solve my problem.

As for your particular problem, it looked to me as though you already solved it. Is that not so?

By the way, the "quote" button allow you to show someone's post as a quote so you don't have to cut and paste it, which gets confusing.
 

FAQ: What Angle Is Needed to Simulate 0.8g Deceleration on an Inclined Plane?

What does it mean to decelerate a body at 0.8g?

Decelerating a body at 0.8g means to apply a force that is equivalent to 0.8 times the force of gravity on Earth (9.8 m/s²). This force causes the body to slow down or come to a stop, depending on its initial velocity.

How is 0.8g deceleration achieved?

0.8g deceleration can be achieved by using various methods such as braking systems, air resistance, or gravitational forces. The specific method used depends on the situation and the type of body being decelerated.

What are the effects of decelerating a body at 0.8g?

The effects of decelerating a body at 0.8g can include a decrease in speed, a change in direction, and an increase in force exerted on the body. This can also cause discomfort or injury to the body if the deceleration is sudden or too intense.

What are the applications of decelerating a body at 0.8g?

Decelerating a body at 0.8g has various applications in different fields such as transportation, sports, and military. It is commonly used in braking systems of vehicles, parachutes for skydiving, and fighter jets for rapid deceleration during combat maneuvers.

How does decelerating a body at 0.8g differ from accelerating it?

Deceleration at 0.8g is the opposite of acceleration at 0.8g. Acceleration increases the body's speed and velocity, while deceleration decreases them. Additionally, deceleration involves a negative change in velocity, whereas acceleration involves a positive change in velocity.

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