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 2 mins ago
Qmechanic♦
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asked 4 hours ago
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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 2 mins ago
Qmechanic♦
101k121821140
asked 4 hours ago
Viktor
161
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
2 hours ago
I vote you are correct and the problem is either misguided or badly worded.
– ja72
11 mins ago
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 2 mins ago
Qmechanic♦
101k121821140
asked 4 hours ago
Viktor
161
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 2 mins ago
Qmechanic♦
101k121821140
asked 4 hours ago
Viktor
161
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 2 mins ago
Qmechanic♦
101k121821140
asked 4 hours ago
Viktor
161
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 2 mins ago
Qmechanic♦
101k121821140
edited 2 mins ago
Qmechanic♦
101k121821140
edited 2 mins ago
Qmechanic♦
101k121821140
101k121821140
asked 4 hours ago
Viktor
161
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 4 hours ago
Viktor
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asked 4 hours ago
Viktor
161
161
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Viktor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Check out our Code of Conduct.
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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
2 hours ago
I vote you are correct and the problem is either misguided or badly worded.
– ja72
11 mins ago
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
2 hours ago
I vote you are correct and the problem is either misguided or badly worded.
– ja72
11 mins ago
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
2 hours ago
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
2 hours ago
I vote you are correct and the problem is either misguided or badly worded.
– ja72
11 mins ago
I vote you are correct and the problem is either misguided or badly worded.
– ja72
11 mins ago
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!
Please also see my comment in the discussion below.
answered 4 hours ago
Philip Wood
7,5323616
Contact force is the vector sum of the frictional force and the normal force so even that is incorrect terminology.
– harshit54
3 hours ago
@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
3 hours ago
Oh, I just saw that they write the formula as $F_n=mg_{eff}$ so that takes the rotation into account. My apologies.
– harshit54
3 hours ago
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
2 hours ago
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
2 hours ago
|
show 1 more 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 4 hours ago
akhmeteli
17.6k21740
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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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!
Please also see my comment in the discussion below.
answered 4 hours ago
Philip Wood
7,5323616
Contact force is the vector sum of the frictional force and the normal force so even that is incorrect terminology.
– harshit54
3 hours ago
@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
3 hours ago
Oh, I just saw that they write the formula as $F_n=mg_{eff}$ so that takes the rotation into account. My apologies.
– harshit54
3 hours ago
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
2 hours ago
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
2 hours ago
|
show 1 more comment
"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!
Please also see my comment in the discussion below.
answered 4 hours ago
Philip Wood
7,5323616
Contact force is the vector sum of the frictional force and the normal force so even that is incorrect terminology.
– harshit54
3 hours ago
@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
3 hours ago
Oh, I just saw that they write the formula as $F_n=mg_{eff}$ so that takes the rotation into account. My apologies.
– harshit54
3 hours ago
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
2 hours ago
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
2 hours ago
|
show 1 more comment
"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!
Please also see my comment in the discussion below.
answered 4 hours ago
Philip Wood
7,5323616
"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!
Please also see my comment in the discussion below.
answered 4 hours ago
Philip Wood
7,5323616
edited 2 hours ago
answered 4 hours ago
Philip Wood
7,5323616
answered 4 hours ago
Philip Wood
7,5323616
answered 4 hours ago
Philip Wood
7,5323616
7,5323616
Contact force is the vector sum of the frictional force and the normal force so even that is incorrect terminology.
– harshit54
3 hours ago
@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
3 hours ago
Oh, I just saw that they write the formula as $F_n=mg_{eff}$ so that takes the rotation into account. My apologies.
– harshit54
3 hours ago
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
2 hours ago
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
2 hours ago
|
show 1 more comment
Contact force is the vector sum of the frictional force and the normal force so even that is incorrect terminology.
– harshit54
3 hours ago
@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
3 hours ago
Oh, I just saw that they write the formula as $F_n=mg_{eff}$ so that takes the rotation into account. My apologies.
– harshit54
3 hours ago
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
2 hours ago
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
2 hours ago
Contact force is the vector sum of the frictional force and the normal force so even that is incorrect terminology.
– harshit54
3 hours ago
Contact force is the vector sum of the frictional force and the normal force so even that is incorrect terminology.
– harshit54
3 hours ago
@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
3 hours ago
@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
3 hours ago
Oh, I just saw that they write the formula as $F_n=mg_{eff}$ so that takes the rotation into account. My apologies.
– harshit54
3 hours ago
Oh, I just saw that they write the formula as $F_n=mg_{eff}$ so that takes the rotation into account. My apologies.
– harshit54
3 hours ago
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
2 hours ago
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
2 hours ago
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
2 hours ago
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
2 hours ago
|
show 1 more 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 4 hours ago
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 4 hours ago
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 4 hours ago
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 4 hours ago
akhmeteli
17.6k21740
answered 4 hours ago
akhmeteli
17.6k21740
answered 4 hours ago
akhmeteli
17.6k21740
answered 4 hours ago
akhmeteli
17.6k21740
17.6k21740
add a comment |
add a comment |
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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
2 hours ago
I vote you are correct and the problem is either misguided or badly worded.
– ja72
11 mins ago