How to Find the Normal Vector for a Particle at an Angle of -30°?

[cloud360]

Homework Statement

Homework Equations

The Attempt at a Solution

If its normal than the doct product is .

i was thinking that i need to show acceleration (dot) velocity=0? is that correct.

If not, what 2 things should i dot?

 

[Mark44]

i was thinking that i need to show acceleration (dot) velocity=0? is that correct.

No. Acceleration and velocity are in the same direction, so their dot product is not zero.

The vectors that are perpendicular are s and n.

To do this problem, you need to resolve the force acting on the mass into its components. One force is the weight of the particle. This force should be resolved into two components - one component is straight into (i.e., perpendicular to) the plane, and the other is parallel to the plane in the same direction as s. The other force is the friction force that is given in the problem. This force is in the opposite direction of s. You are given that the angle is 30 degrees (pi/6). You will need to use this angle in your resolved forces.

You don't need to dot any vectors. If u = <a, b>, the vector v = <-b, a> is perpendicular to u.

 

[cloud360]

No. Acceleration and velocity are in the same direction, so their dot product is not zero.

The vectors that are perpendicular are s and n.

To do this problem, you need to resolve the force acting on the mass into its components. One force is the weight of the particle. This force should be resolved into two components - one component is straight into (i.e., perpendicular to) the plane, and the other is parallel to the plane in the same direction as s. The other force is the friction force that is given in the problem. This force is in the opposite direction of s. You are given that the angle is 30 degrees (pi/6). You will need to use this angle in your resolved forces.

You don't need to dot any vectors. If u = <a, b>, the vector v = <-b, a> is perpendicular to u.

i have problem deriving functions or equation from the given info. please can you kindly tell me how to derive equations for this. or at least give me a place to start with more info than you already have.

When it comes to questions like these i don't know what to do. if there is a technique or "systematic" method of working out an equation from the given info. please tell me

 

[Norfonz]

Be advised that finding a cross product provides a vector normal to those vectors used in the cross product operation.

 

[cloud360]

Be advised that finding a cross product provides a vector normal to those vectors used in the cross product operation.

what are "those vectors"

is it like i(cos(60)) + j(sin(30))+k ?

 

[Mark44]

Be advised that finding a cross product provides a vector normal to those vectors used in the cross product operation.

I don't see that the cross product is relevant to this problem.

 

[Mark44]

i have problem deriving functions or equation from the given info. please can you kindly tell me how to derive equations for this. or at least give me a place to start with more info than you already have.

When it comes to questions like these i don't know what to do. if there is a technique or "systematic" method of working out an equation from the given info. please tell me

You don't need to find an equation. What you need to do is resolve s into its i and j components. All that's involved in doing this is some right triangle trig. Be advised though, that the vertical component has a negative coefficient, and the horizontal component has a positive coefficient. That's because s points down and to the right.

 

[Mark44]

what are "those vectors"

is it like i(cos(60)) + j(sin(30))+k ?

There's no k vector in this problem. Everything is happening in two dimensions.

 

[cloud360]

There's no k vector in this problem. Everything is happening in two dimensions.

so is

icos60 +ksin30

correct? the picture shows a k, am guessing there is no j?

i just guest that. if it is correct can you tell me a systematic method of finding the vectors...please

as i have no idea how to get the answer

 

[Mark44]

i and k are fine and are consistent with the picture, but those look like the x and y axes to me (hence i and j).

so is icos60 +ksin30 correct?

No, but you're on the right track. i cos 30[/color] + k sin 30 would be the components of a unit vector that points up and to the right. s is not a unit vector, and it points down and to the right. How can you get your vector to be the right length and point in the right direction?

Also, sin(30) = 1/2 and cos(30) = sqrt(3)/2

 

[cloud360]

i and k are fine and are consistent with the picture, but those look like the x and y axes to me (hence i and j).No, but you're on the right track. i cos 30[/color] + k sin 30 would be the components of a unit vector that points up and to the right. s is not a unit vector, and it points down and to the right. How can you get your vector to be the right length and point in the right direction?

Also, sin(30) = 1/2 and cos(30) = sqrt(3)/2

q1)so am i correct in saying that the components are always icos(alpha)+isin(alpha) if its 2dor is it icos(90-alpha)+isin(alpha), on certian circumstances. if so which circumstance.

q2) also am i right in saying cos(alpha) represents x-axis usually? and sin(alpha) represents y??

furthermore. i have attempted part iv of this question, see part iv

please can you tell me if my solution to part iv below is correct?

the second line is a re arranged form of F=ma

 

[Mark44]

i cos 30 + k sin 30 would be the components of a unit vector that points up and to the right. s is not a unit vector, and it points down and to the right. How can you get your vector to be the right length and point in the right direction?

so am i correct in saying that the components are always icos(alpha)+isin(alpha) if its 2d

These are the components of a unit vector that makes an angle of alpha with the positive x-axis. You're not dealing with a unit vector. Please reread what I said in my previous post.

or is it icos(90-alpha)+isin(alpha)

No. This is the same as isin(alpha) + isin(alpha) = 2i sin(alpha).

For part iv, I don't know what you got in part iii. I suspect that you aren't solving the right differential equation.

 

[cloud360]

i cos 30 + k sin 30 would be the components of a unit vector that points up and to the right. s is not a unit vector, and it points down and to the right. How can you get your vector to be the right length and point in the right direction?

These are the components of a unit vector that makes an angle of alpha with the positive x-axis. You're not dealing with a unit vector. Please reread what I said in my previous post.

No. This is the same as isin(alpha) + isin(alpha) = 2i sin(alpha).

For part iv, I don't know what you got in part iii. I suspect that you aren't solving the right differential equation.

for part iii, i got the below

 

[Mark44]

for part iii, i got the below

Your differential equation is wrong. Your expression for the force down the plane is ms'', which is incorrect. The downward force on the particle is mg. This force can be (and has to be) resolved into two components, one going down the plane (tangential component) and one acting straight into the plane (the normal component).

You need to make a force resolution diagram of the object. That's what parts i an ii are about.

 

[cloud360]

Your differential equation is wrong. Your expression for the force down the plane is ms'', which is incorrect. The downward force on the particle is mg. This force can be (and has to be) resolved into two components, one going down the plane (tangential component) and one acting straight into the plane (the normal component).

You need to make a force resolution diagram of the object. That's what parts i an ii are about.

i have no idea to do this.

can you please tell me what the differential equation is meant to be.

Or at least tell me how to get the 2 components

 

[Mark44]

i have no idea to do this.

See posts #10 and #12. You were going the right direction, and then you quit.

can you please tell me what the differential equation is meant to be.

Or at least tell me how to get the 2 components

See above.

 

[cloud360]

ok so the second force acting on the body is

F₂=isin30 + kcos30=0.5i+k(sqrt(3)/2))

So the resultant force is

F+F₂=isin30 + kcos30-(a+ns')s^=0.5i+k(sqrt(3)/2))-i(a+ns')=(-(a+ns')+0.5)i+k(sqrt(3)/2))

where s^=s hat=i

Also to confirm. the component when its 2d is usually F=isin(alpha)+jcos(alpha)

where j is the y axis?

because i think i remember seeing in my notes that its isin(alpha)+j(180-alpha), in some cases

i thought it was 90, but now i think it was 180>

 

[Mark44]

ok so the second force acting on the body is

F₂=isin30 + kcos30=0.5i+k(sqrt(3)/2))

No. Apparently you're not understanding what I wrote.
1) These are the compenents of a unit vector. Your s vector is not a unit vector.
2) Besides being a unit vector, your vector above points in the wrong direction. s points down and to the right, so the vertical component should point down (not up). Your horizontal component is pointing inthe right direction.

I said these things before in posts #10 and #12.

So the resultant force is

F+F₂=isin30 + kcos30-(a+ns')s^=0.5i+k(sqrt(3)/2))-i(a+ns')=(-(a+ns')+0.5)i+k(sqrt(3)/2))

where s^=s hat=i

Also to confirm. the component when its 2d is usually F=isin(alpha)+jcos(alpha)

where j is the y axis?

This would be correct if the magnitude of F was 1, and the direction of F made an angle of alpha with the positive x axis.

because i think i remember seeing in my notes that its isin(alpha)+j(180-alpha), in some cases

No, there should be a cosine term. If a vector v makes an angle of θ with the positive x-axis, it can be resolved into two vectors as v = |v|(i cosθ + j sin θ).

i thought it was 90, but now i think it was 180>

There are a lot of different trig identities that can be used.

 

[cloud360]

No. Apparently you're not understanding what I wrote.
1) These are the compenents of a unit vector. Your s vector is not a unit vector.
2) Besides being a unit vector, your vector above points in the wrong direction. s points down and to the right, so the vertical component should point down (not up). Your horizontal component is pointing inthe right direction.

I said these things before in posts #10 and #12.

This would be correct if the magnitude of F was 1, and the direction of F made an angle of alpha with the positive x axis.
No, there should be a cosine term. If a vector v makes an angle of θ with the positive x-axis, it can be resolved into two vectors as v = |v|(i cosθ + j sin θ).
There are a lot of different trig identities that can be used.

My resultant force is wrong? then what is correct resultant force. do i need to multiply F by isin30 +kcos30?

 

[Mark44]

My resultant force is wrong?

Yes, your resultant force is wrong. That's what I said in post #18.

then what is correct resultant force.
do i need to multiply F by isin30 +kcos30?

No. Again, isin30 +kcos30 is not the right unit vector. You need to resolve the downward-acting weight force (mg) into two components, using right triangle trig. One component of the force is the normal force. It acts in a direction perpendicular to the inclined plane. The other component is the tangential force. It acts in a direction down the plane.

You are apparently in a physics class. Your book should have a number of examples showing how to resolve a vector into two perpendicular components.

There are two forces acting on the object:
1) the force due to gravity, mg. This force is directed downward. It needs to be resolved into two components - a normal component, and a tangential component. I think I have said this three times already.
2) the force that resists the sliding motion. It is directed up the ramp. This force is given in your problem.

The resultant force that will be in your differential equation is the vector sum of the tangential force due to gravity (F₁) and the force that resists sliding (F₂).

 

[cloud360]

Yes, your resultant force is wrong. That's what I said in post #18.
No. Again, isin30 +kcos30 is not the right unit vector. You need to resolve the downward-acting weight force (mg) into two components, using right triangle trig. One component of the force is the normal force. It acts in a direction perpendicular to the inclined plane. The other component is the tangential force. It acts in a direction down the plane.

You are apparently in a physics class. Your book should have a number of examples showing how to resolve a vector into two perpendicular components.

There are two forces acting on the object:
1) the force due to gravity, mg. This force is directed downward. It needs to be resolved into two components - a normal component, and a tangential component. I think I have said this three times already.
2) the force that resists the sliding motion. It is directed up the ramp. This force is given in your problem.

The resultant force that will be in your differential equation is the vector sum of the tangential force due to gravity (F₁) and the force that resists sliding (F₂).

my course is calcalus, not physics. it has a mechanics section and i don't understand my lecturers notes.


I learned from this http://en.wikibooks.org/wiki/A-level_Physics/Forces_and_Motion/Scalars_and_vectors

and this
http://www.wikihow.com/Resolve-a-Vector-Into-Components

component1 = magnitude * sin (angle), and

component2 = magnitude * cos (angle).
so

So Fx=|s^|*Cos(30) ,the horizontal component

Fy=|s^|*Sin(30), the vertical component

where s^=s hat

Then Fx+Fy=F₂

i don't know what |s^| is, how do i find it?

 

[HallsofIvy]

Neither of your links has an "s hat" but the first gives Fx= Fcos(A) and Fy= Fsin(A). F is the length of the vector whose components you are trying to find.

 

[cloud360]

Neither of your links has an "s hat" but the first gives Fx= Fcos(A) and Fy= Fsin(A). F is the length of the vector whose components you are trying to find.

and the length is the magnitude so is it not |s^| or is it |s|

 

[Mark44]

my course is calcalus, not physics. it has a mechanics section and i don't understand my lecturers notes.


I learned from this http://en.wikibooks.org/wiki/A-level_Physics/Forces_and_Motion/Scalars_and_vectors

and this
http://www.wikihow.com/Resolve-a-Vector-Into-Components

component1 = magnitude * sin (angle), and

component2 = magnitude * cos (angle).
so

So Fx=|s^|*Cos(30) ,the horizontal component

Fy=|s^|*Sin(30), the vertical component

where s^=s hat

Then Fx+Fy=F₂

i don't know what |s^| is, how do i find it?

Presumably s^ is a unit vector, so |s^| = 1. The magnitude of any unit vector is 1.

Vector s = |s|s^. Any vector is equal to its magnitude times a unit vector in the same direction.

What is s equal to in terms of the weight of the particle? The weight of the particle is a force in the downward direction. The s vector makes an angle of -30°

with the positive x axis. The s vector is the tangential vector I referred to before.