How to calculate the gravity on a hill?

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SUMMARY

This discussion focuses on calculating the gravitational force acting on a box with a mass of 40 kg positioned on a hill with a 25-degree incline. The correct method to find the component of gravitational force acting down the incline is to use the formula Fgravity,x = Fgravity sin(a), resulting in Fgravity,x = 165.8 N. The confusion arose from incorrectly attempting to calculate the hypotenuse using division instead of recognizing that the gravitational force vector acts as the hypotenuse in this scenario. Understanding the relationship between the weight vector and the incline angle is crucial for accurate calculations.

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caspeerrr
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Member advised to use the homework template for posts in the homework sections of PF.
The question is about a box with no movement standing on a hill. The hill has an angle of 25 degrees. The box has a mass of 40 kg.
1. Calculte the gravity
This I still get: F= M x A = 40 x 9,81 = 3,9 x 10^2
The next question tough:
2. Calculate the component Fgravity,x off the gravity alongside the hill.
The component I think they ment is marked on the picture included.

What I did was 390 / sin(25) because the opposite is known (the gravity) and you want to calculate the hypotenuse. The answer indicates this as the right answer tough: Fgravity,x = Fgravity sin(a) = 165,8

Why do they multiply when you should divide?
Am I missing something?
Thanks in advance!
Knipsel.PNG
 
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caspeerrr said:
What I did was 390 / sin(25) because the opposite is known (the gravity) and you want to calculate the hypotenuse.
You want the component of the weight acting down the incline. When finding the component of a vector using a right triangle, the full vector is always the hypotenuse. (The components are always smaller than the full vector.)

You're using the wrong triangle: Draw the weight vector acting straight down. That's the hypotenuse of the correct triangle.
 
Doc Al said:
You want the component of the weight acting down the incline. When finding the component of a vector using a right triangle, the full vector is always the hypotenuse. (The components are always smaller than the full vector.)

You're using the wrong triangle: Draw the weight vector acting straight down. That's the hypotenuse of the correct triangle.
Doc Al said:
You want the component of the weight acting down the incline. When finding the component of a vector using a right triangle, the full vector is always the hypotenuse. (The components are always smaller than the full vector.)

You're using the wrong triangle: Draw the weight vector acting straight down. That's the hypotenuse of the correct triangle.

Aaah I get it, thank you!
 
If you find yourself in an exam and can't remember if it's sin or cos... consider what happens if the angle of the slope is reduced towards zero degrees (no slope). The force you calculate down the slope should approach zero. Your calculator will tell you sin(0)=0 but cos(0)=1.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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