Normal force not perpendicular to the surface
In my class about mechanics i had to solve this problem, but it was never really explained. The solution is found beneath in a picture. In the solution they also calculate the angle of the normal force, but isn't the normal force always perpendicular to the contact surface in the contact point, so why isn't the normal force facing outwards and perpendicular to the tangent in that point. I've asked a lot of other students in my class, and none of them seem to understand this.
newtonian-mechanics forces terminology vectors centripetal-force
edited Dec 22 at 20:52
Qmechanic♦
101k121831142
asked Dec 22 at 16:23
Viktor
233
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add a comment |
In my class about mechanics i had to solve this problem, but it was never really explained. The solution is found beneath in a picture. In the solution they also calculate the angle of the normal force, but isn't the normal force always perpendicular to the contact surface in the contact point, so why isn't the normal force facing outwards and perpendicular to the tangent in that point. I've asked a lot of other students in my class, and none of them seem to understand this.
newtonian-mechanics forces terminology vectors centripetal-force
edited Dec 22 at 20:52
Qmechanic♦
101k121831142
asked Dec 22 at 16:23
Viktor
233
New contributor
Viktor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Usually the normal force is points perpendicular outwards from the surface like you said, but in this case in makes sense to define the normal force in a special way. Read the definition closely. (Maybe you think this definition doesn't make sense. You can think that, but as a good physicist you should still be able to solve the problem.)
– psitae
Dec 22 at 18:37
I vote you are correct and the problem is either misguided or badly worded.
– ja72
Dec 22 at 20:43
add a comment |
In my class about mechanics i had to solve this problem, but it was never really explained. The solution is found beneath in a picture. In the solution they also calculate the angle of the normal force, but isn't the normal force always perpendicular to the contact surface in the contact point, so why isn't the normal force facing outwards and perpendicular to the tangent in that point. I've asked a lot of other students in my class, and none of them seem to understand this.
newtonian-mechanics forces terminology vectors centripetal-force
edited Dec 22 at 20:52
Qmechanic♦
101k121831142
asked Dec 22 at 16:23
Viktor
233
New contributor
Viktor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In my class about mechanics i had to solve this problem, but it was never really explained. The solution is found beneath in a picture. In the solution they also calculate the angle of the normal force, but isn't the normal force always perpendicular to the contact surface in the contact point, so why isn't the normal force facing outwards and perpendicular to the tangent in that point. I've asked a lot of other students in my class, and none of them seem to understand this.
newtonian-mechanics forces terminology vectors centripetal-force
newtonian-mechanics forces terminology vectors centripetal-force
edited Dec 22 at 20:52
Qmechanic♦
101k121831142
asked Dec 22 at 16:23
Viktor
233
New contributor
Viktor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 22 at 20:52
Qmechanic♦
101k121831142
asked Dec 22 at 16:23
Viktor
233
New contributor
Viktor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 22 at 20:52
Qmechanic♦
101k121831142
edited Dec 22 at 20:52
Qmechanic♦
101k121831142
edited Dec 22 at 20:52
Qmechanic♦
101k121831142
101k121831142
asked Dec 22 at 16:23
Viktor
233
New contributor
Viktor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 22 at 16:23
Viktor
233
asked Dec 22 at 16:23
Viktor
233
233
New contributor
Viktor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Viktor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Viktor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Usually the normal force is points perpendicular outwards from the surface like you said, but in this case in makes sense to define the normal force in a special way. Read the definition closely. (Maybe you think this definition doesn't make sense. You can think that, but as a good physicist you should still be able to solve the problem.)
– psitae
Dec 22 at 18:37
I vote you are correct and the problem is either misguided or badly worded.
– ja72
Dec 22 at 20:43
add a comment |
Usually the normal force is points perpendicular outwards from the surface like you said, but in this case in makes sense to define the normal force in a special way. Read the definition closely. (Maybe you think this definition doesn't make sense. You can think that, but as a good physicist you should still be able to solve the problem.)
– psitae
Dec 22 at 18:37
I vote you are correct and the problem is either misguided or badly worded.
– ja72
Dec 22 at 20:43
Usually the normal force is points perpendicular outwards from the surface like you said, but in this case in makes sense to define the normal force in a special way. Read the definition closely. (Maybe you think this definition doesn't make sense. You can think that, but as a good physicist you should still be able to solve the problem.)
– psitae
Dec 22 at 18:37
Usually the normal force is points perpendicular outwards from the surface like you said, but in this case in makes sense to define the normal force in a special way. Read the definition closely. (Maybe you think this definition doesn't make sense. You can think that, but as a good physicist you should still be able to solve the problem.)
– psitae
Dec 22 at 18:37
I vote you are correct and the problem is either misguided or badly worded.
– ja72
Dec 22 at 20:43
I vote you are correct and the problem is either misguided or badly worded.
– ja72
Dec 22 at 20:43
add a comment |
2 Answers
2
active
oldest
votes
"Normal force" is simply bad notation. "Contact force" would be better. Usually, we don't distinguish, because the contact force is almost normal to the surface. But in the context of this detailed examination of the rotating Earth, it is confusing not to distinguish!
Later Additions (incorporating comments)
The contact force can be resolved into a component normal to the Earth (modelled as a sphere) and a small tangential (or frictional) component. If this component wasn't present, the body would be slipping round the Earth's surface, towards the equator!
I can't resist remarking that the 'textbook' treatment reproduced in the question is terribly long-winded. The results can be obtained in three or four lines by applying the cosine formula and the sine formula to a simple vector triangle.
answered Dec 22 at 16:41
Philip Wood
7,6073616
1
Contact force is the vector sum of the frictional force and the normal force so even that is incorrect terminology.
– harshit54
Dec 22 at 17:04
1
@Harshit54 "Contact force is the vector sum of the frictional force and the normal force." Agreed, but that's why the terminology is right! The force $mathbf{F_{N}}$ is the contact force and can indeed be split into normal and tangential components if required. You can call the tangential force the frictional component of the contact force if you wish. If this component wasn't present, the body would be slipping round the Earth's surface, towards the equator!
– Philip Wood
Dec 22 at 17:16
1
Oh, I just saw that they write the formula as $F_n=mg_{eff}$ so that takes the rotation into account. My apologies.
– harshit54
Dec 22 at 17:19
1
Agreed. The term "normal" means perpendicular or orthogonal to a surface. If a force isn't perpendicular to a surface, then it isn't a normal force, by definition.
– The Photon
Dec 22 at 18:22
1
Maybe the textbook author was trying to give an example of using $textbf{i}$ and $textbf j$ notation, but his/her solution is terribly long-winded. The whole thing can be done in 3 or 4 lines by applying sine and cosine formulae to a simple vector triangle.
– Philip Wood
Dec 22 at 18:27
|
show 3 more comments
I think they just use incorrect terminology. What they call "normal force" is actually a vector sum of the normal force and the force of friction (the body would not rest in an arbitrary point of the surface of rotating Earth without friction).
answered Dec 22 at 16:44
akhmeteli
17.6k21740
add a comment |
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2 Answers
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active
oldest
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2 Answers
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oldest
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"Normal force" is simply bad notation. "Contact force" would be better. Usually, we don't distinguish, because the contact force is almost normal to the surface. But in the context of this detailed examination of the rotating Earth, it is confusing not to distinguish!
Later Additions (incorporating comments)
The contact force can be resolved into a component normal to the Earth (modelled as a sphere) and a small tangential (or frictional) component. If this component wasn't present, the body would be slipping round the Earth's surface, towards the equator!
I can't resist remarking that the 'textbook' treatment reproduced in the question is terribly long-winded. The results can be obtained in three or four lines by applying the cosine formula and the sine formula to a simple vector triangle.
answered Dec 22 at 16:41
Philip Wood
7,6073616
1
Contact force is the vector sum of the frictional force and the normal force so even that is incorrect terminology.
– harshit54
Dec 22 at 17:04
1
@Harshit54 "Contact force is the vector sum of the frictional force and the normal force." Agreed, but that's why the terminology is right! The force $mathbf{F_{N}}$ is the contact force and can indeed be split into normal and tangential components if required. You can call the tangential force the frictional component of the contact force if you wish. If this component wasn't present, the body would be slipping round the Earth's surface, towards the equator!
– Philip Wood
Dec 22 at 17:16
1
Oh, I just saw that they write the formula as $F_n=mg_{eff}$ so that takes the rotation into account. My apologies.
– harshit54
Dec 22 at 17:19
1
Agreed. The term "normal" means perpendicular or orthogonal to a surface. If a force isn't perpendicular to a surface, then it isn't a normal force, by definition.
– The Photon
Dec 22 at 18:22
1
Maybe the textbook author was trying to give an example of using $textbf{i}$ and $textbf j$ notation, but his/her solution is terribly long-winded. The whole thing can be done in 3 or 4 lines by applying sine and cosine formulae to a simple vector triangle.
– Philip Wood
Dec 22 at 18:27
|
show 3 more comments
"Normal force" is simply bad notation. "Contact force" would be better. Usually, we don't distinguish, because the contact force is almost normal to the surface. But in the context of this detailed examination of the rotating Earth, it is confusing not to distinguish!
Later Additions (incorporating comments)
The contact force can be resolved into a component normal to the Earth (modelled as a sphere) and a small tangential (or frictional) component. If this component wasn't present, the body would be slipping round the Earth's surface, towards the equator!
I can't resist remarking that the 'textbook' treatment reproduced in the question is terribly long-winded. The results can be obtained in three or four lines by applying the cosine formula and the sine formula to a simple vector triangle.
answered Dec 22 at 16:41
Philip Wood
7,6073616
1
Contact force is the vector sum of the frictional force and the normal force so even that is incorrect terminology.
– harshit54
Dec 22 at 17:04
1
@Harshit54 "Contact force is the vector sum of the frictional force and the normal force." Agreed, but that's why the terminology is right! The force $mathbf{F_{N}}$ is the contact force and can indeed be split into normal and tangential components if required. You can call the tangential force the frictional component of the contact force if you wish. If this component wasn't present, the body would be slipping round the Earth's surface, towards the equator!
– Philip Wood
Dec 22 at 17:16
1
Oh, I just saw that they write the formula as $F_n=mg_{eff}$ so that takes the rotation into account. My apologies.
– harshit54
Dec 22 at 17:19
1
Agreed. The term "normal" means perpendicular or orthogonal to a surface. If a force isn't perpendicular to a surface, then it isn't a normal force, by definition.
– The Photon
Dec 22 at 18:22
1
Maybe the textbook author was trying to give an example of using $textbf{i}$ and $textbf j$ notation, but his/her solution is terribly long-winded. The whole thing can be done in 3 or 4 lines by applying sine and cosine formulae to a simple vector triangle.
– Philip Wood
Dec 22 at 18:27
|
show 3 more comments
"Normal force" is simply bad notation. "Contact force" would be better. Usually, we don't distinguish, because the contact force is almost normal to the surface. But in the context of this detailed examination of the rotating Earth, it is confusing not to distinguish!
Later Additions (incorporating comments)
The contact force can be resolved into a component normal to the Earth (modelled as a sphere) and a small tangential (or frictional) component. If this component wasn't present, the body would be slipping round the Earth's surface, towards the equator!
I can't resist remarking that the 'textbook' treatment reproduced in the question is terribly long-winded. The results can be obtained in three or four lines by applying the cosine formula and the sine formula to a simple vector triangle.
answered Dec 22 at 16:41
Philip Wood
7,6073616
"Normal force" is simply bad notation. "Contact force" would be better. Usually, we don't distinguish, because the contact force is almost normal to the surface. But in the context of this detailed examination of the rotating Earth, it is confusing not to distinguish!
Later Additions (incorporating comments)
The contact force can be resolved into a component normal to the Earth (modelled as a sphere) and a small tangential (or frictional) component. If this component wasn't present, the body would be slipping round the Earth's surface, towards the equator!
I can't resist remarking that the 'textbook' treatment reproduced in the question is terribly long-winded. The results can be obtained in three or four lines by applying the cosine formula and the sine formula to a simple vector triangle.
answered Dec 22 at 16:41
Philip Wood
7,6073616
edited 2 days ago
answered Dec 22 at 16:41
Philip Wood
7,6073616
answered Dec 22 at 16:41
Philip Wood
7,6073616
answered Dec 22 at 16:41
Philip Wood
7,6073616
7,6073616
1
Contact force is the vector sum of the frictional force and the normal force so even that is incorrect terminology.
– harshit54
Dec 22 at 17:04
1
@Harshit54 "Contact force is the vector sum of the frictional force and the normal force." Agreed, but that's why the terminology is right! The force $mathbf{F_{N}}$ is the contact force and can indeed be split into normal and tangential components if required. You can call the tangential force the frictional component of the contact force if you wish. If this component wasn't present, the body would be slipping round the Earth's surface, towards the equator!
– Philip Wood
Dec 22 at 17:16
1
Oh, I just saw that they write the formula as $F_n=mg_{eff}$ so that takes the rotation into account. My apologies.
– harshit54
Dec 22 at 17:19
1
Agreed. The term "normal" means perpendicular or orthogonal to a surface. If a force isn't perpendicular to a surface, then it isn't a normal force, by definition.
– The Photon
Dec 22 at 18:22
1
Maybe the textbook author was trying to give an example of using $textbf{i}$ and $textbf j$ notation, but his/her solution is terribly long-winded. The whole thing can be done in 3 or 4 lines by applying sine and cosine formulae to a simple vector triangle.
– Philip Wood
Dec 22 at 18:27
|
show 3 more comments
1
Contact force is the vector sum of the frictional force and the normal force so even that is incorrect terminology.
– harshit54
Dec 22 at 17:04
1
@Harshit54 "Contact force is the vector sum of the frictional force and the normal force." Agreed, but that's why the terminology is right! The force $mathbf{F_{N}}$ is the contact force and can indeed be split into normal and tangential components if required. You can call the tangential force the frictional component of the contact force if you wish. If this component wasn't present, the body would be slipping round the Earth's surface, towards the equator!
– Philip Wood
Dec 22 at 17:16
1
Oh, I just saw that they write the formula as $F_n=mg_{eff}$ so that takes the rotation into account. My apologies.
– harshit54
Dec 22 at 17:19
1
Agreed. The term "normal" means perpendicular or orthogonal to a surface. If a force isn't perpendicular to a surface, then it isn't a normal force, by definition.
– The Photon
Dec 22 at 18:22
1
Maybe the textbook author was trying to give an example of using $textbf{i}$ and $textbf j$ notation, but his/her solution is terribly long-winded. The whole thing can be done in 3 or 4 lines by applying sine and cosine formulae to a simple vector triangle.
– Philip Wood
Dec 22 at 18:27
1
1
Contact force is the vector sum of the frictional force and the normal force so even that is incorrect terminology.
– harshit54
Dec 22 at 17:04
Contact force is the vector sum of the frictional force and the normal force so even that is incorrect terminology.
– harshit54
Dec 22 at 17:04
1
1
@Harshit54 "Contact force is the vector sum of the frictional force and the normal force." Agreed, but that's why the terminology is right! The force $mathbf{F_{N}}$ is the contact force and can indeed be split into normal and tangential components if required. You can call the tangential force the frictional component of the contact force if you wish. If this component wasn't present, the body would be slipping round the Earth's surface, towards the equator!
– Philip Wood
Dec 22 at 17:16
@Harshit54 "Contact force is the vector sum of the frictional force and the normal force." Agreed, but that's why the terminology is right! The force $mathbf{F_{N}}$ is the contact force and can indeed be split into normal and tangential components if required. You can call the tangential force the frictional component of the contact force if you wish. If this component wasn't present, the body would be slipping round the Earth's surface, towards the equator!
– Philip Wood
Dec 22 at 17:16
1
1
Oh, I just saw that they write the formula as $F_n=mg_{eff}$ so that takes the rotation into account. My apologies.
– harshit54
Dec 22 at 17:19
Oh, I just saw that they write the formula as $F_n=mg_{eff}$ so that takes the rotation into account. My apologies.
– harshit54
Dec 22 at 17:19
1
1
Agreed. The term "normal" means perpendicular or orthogonal to a surface. If a force isn't perpendicular to a surface, then it isn't a normal force, by definition.
– The Photon
Dec 22 at 18:22
Agreed. The term "normal" means perpendicular or orthogonal to a surface. If a force isn't perpendicular to a surface, then it isn't a normal force, by definition.
– The Photon
Dec 22 at 18:22
1
1
Maybe the textbook author was trying to give an example of using $textbf{i}$ and $textbf j$ notation, but his/her solution is terribly long-winded. The whole thing can be done in 3 or 4 lines by applying sine and cosine formulae to a simple vector triangle.
– Philip Wood
Dec 22 at 18:27
Maybe the textbook author was trying to give an example of using $textbf{i}$ and $textbf j$ notation, but his/her solution is terribly long-winded. The whole thing can be done in 3 or 4 lines by applying sine and cosine formulae to a simple vector triangle.
– Philip Wood
Dec 22 at 18:27
|
show 3 more comments
I think they just use incorrect terminology. What they call "normal force" is actually a vector sum of the normal force and the force of friction (the body would not rest in an arbitrary point of the surface of rotating Earth without friction).
answered Dec 22 at 16:44
akhmeteli
17.6k21740
add a comment |
I think they just use incorrect terminology. What they call "normal force" is actually a vector sum of the normal force and the force of friction (the body would not rest in an arbitrary point of the surface of rotating Earth without friction).
answered Dec 22 at 16:44
akhmeteli
17.6k21740
add a comment |
I think they just use incorrect terminology. What they call "normal force" is actually a vector sum of the normal force and the force of friction (the body would not rest in an arbitrary point of the surface of rotating Earth without friction).
answered Dec 22 at 16:44
akhmeteli
17.6k21740
I think they just use incorrect terminology. What they call "normal force" is actually a vector sum of the normal force and the force of friction (the body would not rest in an arbitrary point of the surface of rotating Earth without friction).
answered Dec 22 at 16:44
akhmeteli
17.6k21740
answered Dec 22 at 16:44
akhmeteli
17.6k21740
answered Dec 22 at 16:44
akhmeteli
17.6k21740
answered Dec 22 at 16:44
akhmeteli
17.6k21740
17.6k21740
add a comment |
add a comment |
Viktor is a new contributor. Be nice, and check out our Code of Conduct.
Viktor is a new contributor. Be nice, and check out our Code of Conduct.
Viktor is a new contributor. Be nice, and check out our Code of Conduct.
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Usually the normal force is points perpendicular outwards from the surface like you said, but in this case in makes sense to define the normal force in a special way. Read the definition closely. (Maybe you think this definition doesn't make sense. You can think that, but as a good physicist you should still be able to solve the problem.)
– psitae
Dec 22 at 18:37
I vote you are correct and the problem is either misguided or badly worded.
– ja72
Dec 22 at 20:43