Calculating the Component of the weight that acts along a line

Click For Summary

Homework Help Overview

The problem involves a cyclist riding up an incline and requires calculating the component of the weight acting along that incline. The context is rooted in physics, specifically in the study of forces and inclines.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of trigonometric functions to resolve the weight vector into components. There is uncertainty about the correct application of the sine function and the interpretation of the mark scheme. Some participants attempt to calculate the weight component using the angle derived from the incline.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and questioning the calculations. Some guidance has been offered regarding the importance of sketching the problem and resolving the weight vector, but there is no consensus on the correct approach or final answer.

Contextual Notes

Participants express confusion over the relationship between the weight, mass, and gravitational force, indicating a need for clarification on these fundamental concepts. The problem constraints include neglecting frictional forces and focusing solely on the weight component along the incline.

JudgeA
Messages
8
Reaction score
1

Homework Statement


A cyclist rides along a road up an incline at a steady speed of 9.0 m s–1. The mass of the rider and bicycle is 70kg and the bicycle travels 15 m along the road for every 1.0 m gained in height. Neglect energy loss due to frictional forces.

Calculate the component of the weight of the bicycle and the rider that acts along the incline.

2. Homework Equations

sinθ=opp/hyp, not sure what else I need to use

The Attempt at a Solution



Apparently the answer is 46N but I'm really not sure how to get to that. I found an angle of 3.8 degrees by doing 1.0/15 but I'm not sure where to go from there. The Mark scheme says F=sinθ=Wx1.0/15 I don't quite understand what it's saying.

Thanks
 
Physics news on Phys.org
Try drawing a diagram showing the weight vector. Then resolve the weight vector into two components, one parallel with the incline and one perpendicular to the incline.
 
JudgeA said:
The Mark scheme says F=sinθ=Wx1.0/15 I don't quite understand what it's saying.

I think that's wrong. It should say..

F = W * sin(θ) = W * 1.0/15
 
  • Like
Likes   Reactions: Carol Sanders
CWatters said:
Try drawing a diagram showing the weight vector. Then resolve the weight vector into two components, one parallel with the incline and one perpendicular to the incline.

Yeah I did try that but ended up with 4.6 not 46. I did Sin3.8*70 and got 4.6 so I'm not sure if my angle is wrong?
 
CWatters said:
I think that's wrong. It should say..

F = W * sin(θ) = W * 1.0/15
Ah ok thanks
 
I think every incline problem should start with a sketch, something like the attached file. 3.8 degrees is correct and you found that by knowing the sine of the angle is the sin = opp/hyp = 1.0/15. Referring to the attached pic, if you know W, (the hypotenuse) how do you calculate the opposite side (assuming phi is the angle) of the triangle?
 

Attachments

  • Bike.PNG
    Bike.PNG
    2.7 KB · Views: 619
Vector1962 said:
I think every incline problem should start with a sketch, something like the attached file. 3.8 degrees is correct and you found that by knowing the sine of the angle is the sin = opp/hyp = 1.0/15. Referring to the attached pic, if you know W, (the hypotenuse) how do you calculate the opposite side (assuming phi is the angle) of the triangle?
Vector1962 said:
I think every incline problem should start with a sketch, something like the attached file. 3.8 degrees is correct and you found that by knowing the sine of the angle is the sin = opp/hyp = 1.0/15. Referring to the attached pic, if you know W, (the hypotenuse) how do you calculate the opposite side (assuming phi is the angle) of the triangle?
I assume you'd do Sin3.8*70 as in Opp=Sinθ*hyp however that left me with the answer 4.6 not 46.
(thanks for helping btw)
 
JudgeA said:
I assume you'd do Sin3.8*70 as in Opp=Sinθ*hyp however that left me with the answer 4.6 not 46.
(thanks for helping btw)
What does the force, W equal in terms of the mass, m? W=m x _____
 
JudgeA said:
I assume you'd do Sin3.8*70 as in Opp=Sinθ*hyp however that left me with the answer 4.6 not 46
Vector1962 said:
What does the force, W equal in terms of the mass, m? W=m x _____
Oh of course mass doesn't taken into account gravity, I understand now. Thank you!
 
  • Like
Likes   Reactions: CWatters

Similar threads

  • · Replies 9 ·
Replies
9
Views
6K
Replies
6
Views
3K
  • · Replies 5 ·
Replies
5
Views
8K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
7
Views
7K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
24
Views
12K
  • · Replies 15 ·
Replies
15
Views
5K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
6K