Resolution of vectors for problems related to mechanics

• ubergewehr273
In summary: In general it doesn't matter, but often one of the reference frames is more advantageous (for the calculation and the illustration of the results).
ubergewehr273

Homework Statement

Refer the given image. [Prob 2.9]

F=ma

The Attempt at a Solution

I drew the normal vector perpendicular to the surface of the cone and resolved it as
##Nsin\theta=mg##
##Ncos\theta=\frac {mv_{0}^2} {r}## where ##v_{0}## and ##r## are the speed and radius of the trajectory respectively.
Solving for ##r## we get,
##r=\frac {v_{0}^2} {g}tan\theta##

My doubt is however is on the matter of resolving the required vectors. Is it necessary to resolve the normal reaction into components or is it correct to resolve ##mg## along the normal and along the lateral surface of the cone? Similar situations arise when solving problems related to pendulums. I get confused whether tension vector must be resolved or mg.

Attachments

• 2018-08-17 11.11.42.png
25.5 KB · Views: 421
Last edited:
It does not matter. It is just a question of which basis you work in.

The point if you resolve the normal and the lateral is that the acceleration also needs to be resolved into those components.

ubergewehr273 said:

The Attempt at a Solution

I drew the normal vector perpendicular to the surface of the cone and resolved it as
##Nsin\theta=mg##
##Ncos\theta=\frac {mv_{0}^2} {r}## where ##v_{0}## and ##r## are the speed and radius of the trajectory respectively.
Solving for ##r## we get,
##r=\frac {v_{0}^2} {g}cos\theta##

Unfortunately there is something wrong with the math - I think the approach is correct.

ubergewehr273 said:

The Attempt at a Solution

My doubt is however is on the matter of resolving the required vectors. Is it necessary to resolve the normal reaction into components or is it correct to resolve mg along the normal and along the lateral surface of the cone? Similar situations arise when solving problems related to pendulums. I get confused whether tension vector must be resolved or mg.

In general it doesn't matter, but often one of the reference frames is more advantageous (for the calculation and the illustration of the results).

stockzahn said:
Unfortunately there is something wrong with the math - I think the approach is correct.
I agree. The algebra done to get from the resolved components to the final result is not correct.

ubergewehr273 said:

Homework Statement

Refer the given image. [Prob 2.9]

F=ma

The Attempt at a Solution

I drew the normal vector perpendicular to the surface of the cone and resolved it as
##Nsin\theta=mg##
##Ncos\theta=\frac {mv_{0}^2} {r}## where ##v_{0}## and ##r## are the speed and radius of the trajectory respectively.
Correct

ubergewehr273 said:
Solving for ##r## we get,
##r=\frac {v_{0}^2} {g}cos\theta##
No, it is not right!

I'm apologize for my ignorance. I've made the required change.
Orodruin said:
I agree. The algebra done to get from the resolved components to the final result is not correct.

What's wrong in this approach ? (refer the image provided)

Attachments

• 2018-08-17 15.51.11.jpg
19.4 KB · Views: 364
Also if any of the vectors can be resolved then when resolving acceleration and mg in the direction perpendicular to the surface and along it then we get...
(refer image).
However the body doesn't move along the surface but through this approach a net force appears. Where am I going wrong?

Attachments

• 2018-08-17 15.47.32.jpg
26 KB · Views: 376
ubergewehr273 said:
What's wrong in this approach ? (refer the image provided)

The approach is correct:

ubergewehr273 said:
##Nsin\theta=mg##
##Ncos\theta=\frac {mv_{0}^2} {r}## where ##v_{0}## and ##r## are the speed and radius of the trajectory respectively.

The final result (you showed in your original first post) was not though (like ehild already pointed out):

ubergewehr273 said:
Solving for ##r## we get,
##r=\frac {v_{0}^2} {g}cos\theta##

You seemed to change the ##cos## into a ##tan## afterwards and it now is right.

ubergewehr273 said:
Also if any of the vectors can be resolved then when resolving acceleration and mg in the direction perpendicular to the surface and along it then we get...
(refer image).
However the body doesn't move along the surface but through this approach a net force appears. Where am I going wrong?

In your calculation shown in the pic, the signs are not correct. The component of gravitational force points downwards, the component of the centrigugal force upwards. If you change the sign from plus to minus (and set the result zero) you will obtain the same correlation as in your first post.

stockzahn said:
In your calculation shown in the pic, the signs are not correct. The component of gravitational force points downwards, the component of the centrigugal force upwards. If you change the sign from plus to minus (and set the result zero) you will obtain the same correlation as in your first post.
Thanks but another doubt arises in my mind. When am I supposed to consider centripetal acceleration (towards the centre of trajectory) and centrifugal acceleration (away from centre) ?

stockzahn said:
The approach is correct:
Let's say the problem was to find the magnitude of normal reaction acting on the body. Clearly both the values of normal can't be the same (refer 2nd image). Then what's the right value of the normal reaction ?

ubergewehr273 said:
Thanks but another doubt arises in my mind. When am I supposed to consider centripetal acceleration (towards the centre of trajectory) and centrifugal acceleration (away from centre) ?

There is an elementary difference between the centripetal force, which is a "real" force, and the centrifugal force, which is an inertial, and therefore, a pseudo force. You can compare it with the application of Newton's second law:

$$m a = F_{net}$$

The net force ##F_{net}## corresponds to the real force acting on the ball by the wall (and the gravity) - the centripetal force. The LHS corresponds to the centrifugal force, which is the the resulting change of kinematic state due to the centripetal force, but since they are equal you can take advantage of the formula for the centrifugal force for movements on circular trajectories.

ubergewehr273 said:
Let's say the problem was to find the magnitude of normal reaction acting on the body. Clearly both the values of normal can't be the same (refer 2nd image). Then what's the right value of the normal reaction ?

I'm not sure if I misunderstand, do you doubt that ##N=\frac{mg}{sin\theta}=\frac{mv^2}{r\,cos\theta}## (if the height of the ball is constant)?

stockzahn said:
I'm not sure if I misunderstand, do you doubt that ##N=\frac{mg}{sin\theta}=\frac{mv^2}{r\,cos\theta}## (if the height of the ball is constant)?
Let me start from scratch.
Case 1: I resolve normal reaction vector as one along y-axis as ##Nsin\theta=mg##
Case 2: I resolve ##mg## along the surface of cone and perpendicular to the surface as ##mgsin\theta=N##
Now by your previous statement both the methods are correct however we end up with 2 different values of ##N##. This is what I'm confused about.

ubergewehr273 said:
Let me start from scratch.
Case 1: I resolve normal reaction vector as one along y-axis as ##Nsin\theta=mg##
Case 2: I resolve ##mg## along the surface of cone and perpendicular to the surface as ##mgsin\theta=N##
Now by your previous statement both the methods are correct however we end up with 2 different values of ##N##. This is what I'm confused about.
Your case 2 is wrong. You forgot that the total component perpendicular to the surface needs to be non-zero because the acceleration has a component that is perpendicular to the surface. The correct equation is
$$N - mg\sin(\theta) = \frac{mv_0^2}{r} \cos(\theta).$$

ubergewehr273 said:
Let me start from scratch.
Case 1: I resolve normal reaction vector as one along y-axis as ##Nsin\theta=mg##
Case 2: I resolve ##mg## along the surface of cone and perpendicular to the surface as ##mgsin\theta=N##
Now by your previous statement both the methods are correct however we end up with 2 different values of ##N##. This is what I'm confused about.

Attachments

• BallCone_01.jpg
22.1 KB · Views: 348

1. What is the concept of vector resolution in mechanics?

The concept of vector resolution in mechanics involves breaking down a vector into its components in order to analyze its effect on an object's motion. This is often used in problems related to forces and motion, where different forces act on an object in different directions.

2. How do you resolve a vector into its components?

To resolve a vector into its components, you can use trigonometric functions such as sine, cosine, and tangent. The magnitude of the components can be calculated using the length of the vector and the angle it makes with the x and y axes. The direction of the components can be determined using the angle and the direction of the original vector.

3. What is the difference between scalar and vector resolution?

Scalar resolution involves breaking down a vector into its magnitude or numerical value, while vector resolution involves breaking down a vector into its components in terms of direction and magnitude. Scalar resolution is used when the direction of the vector is not needed for the problem, while vector resolution is used when the direction is important in determining the motion of an object.

4. Can you use vector resolution for non-linear motion problems?

Yes, vector resolution can be used for non-linear motion problems. In such cases, the components of the vector will change as the object moves along its path. The components can be calculated at different points along the path to analyze the overall effect of the vector on the object's motion.

5. How does vector resolution help in solving mechanics problems?

Vector resolution helps in solving mechanics problems by simplifying complex vectors into their components, which are easier to analyze and manipulate mathematically. It also allows for the application of trigonometric principles to determine the overall effect of multiple vectors acting on an object, making it a useful tool for solving problems related to forces and motion.

• Introductory Physics Homework Help
Replies
7
Views
354
• Introductory Physics Homework Help
Replies
2
Views
482
• Introductory Physics Homework Help
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
992
• Introductory Physics Homework Help
Replies
27
Views
3K
• Introductory Physics Homework Help
Replies
9
Views
967
• Introductory Physics Homework Help
Replies
14
Views
3K
• Introductory Physics Homework Help
Replies
2
Views
840
• Introductory Physics Homework Help
Replies
10
Views
861
• Introductory Physics Homework Help
Replies
5
Views
870