Normal force not perpendicular to the surface
$begingroup$
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 '18 at 20:52
Qmechanic♦
103k121851173
asked Dec 22 '18 at 16:23
ViktorViktor
234
$endgroup$
add a comment |
$begingroup$
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 '18 at 20:52
Qmechanic♦
103k121851173
asked Dec 22 '18 at 16:23
ViktorViktor
234
$endgroup$
$begingroup$
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.)
$endgroup$
– psitae
Dec 22 '18 at 18:37
$begingroup$
I vote you are correct and the problem is either misguided or badly worded.
$endgroup$
– ja72
Dec 22 '18 at 20:43
add a comment |
$begingroup$
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 '18 at 20:52
Qmechanic♦
103k121851173
asked Dec 22 '18 at 16:23
ViktorViktor
234
$endgroup$
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 '18 at 20:52
Qmechanic♦
103k121851173
asked Dec 22 '18 at 16:23
ViktorViktor
234
edited Dec 22 '18 at 20:52
Qmechanic♦
103k121851173
asked Dec 22 '18 at 16:23
ViktorViktor
234
edited Dec 22 '18 at 20:52
Qmechanic♦
103k121851173
edited Dec 22 '18 at 20:52
Qmechanic♦
103k121851173
edited Dec 22 '18 at 20:52
Qmechanic♦
103k121851173
103k121851173
asked Dec 22 '18 at 16:23
ViktorViktor
234
asked Dec 22 '18 at 16:23
ViktorViktor
234
asked Dec 22 '18 at 16:23
ViktorViktor
234
234
$begingroup$
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.)
$endgroup$
– psitae
Dec 22 '18 at 18:37
$begingroup$
I vote you are correct and the problem is either misguided or badly worded.
$endgroup$
– ja72
Dec 22 '18 at 20:43
add a comment |
$begingroup$
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.)
$endgroup$
– psitae
Dec 22 '18 at 18:37
$begingroup$
I vote you are correct and the problem is either misguided or badly worded.
$endgroup$
– ja72
Dec 22 '18 at 20:43
$begingroup$
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.)
$endgroup$
– psitae
Dec 22 '18 at 18:37
$begingroup$
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.)
$endgroup$
– psitae
Dec 22 '18 at 18:37
$begingroup$
I vote you are correct and the problem is either misguided or badly worded.
$endgroup$
– ja72
Dec 22 '18 at 20:43
$begingroup$
I vote you are correct and the problem is either misguided or badly worded.
$endgroup$
– ja72
Dec 22 '18 at 20:43
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
"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 '18 at 16:41
Philip WoodPhilip Wood
7,9733616
$endgroup$
1
$begingroup$
Contact force is the vector sum of the frictional force and the normal force so even that is incorrect terminology.
$endgroup$
– harshit54
Dec 22 '18 at 17:04
1
$begingroup$
@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!
$endgroup$
– Philip Wood
Dec 22 '18 at 17:16
1
$begingroup$
Oh, I just saw that they write the formula as $F_n=mg_{eff}$ so that takes the rotation into account. My apologies.
$endgroup$
– harshit54
Dec 22 '18 at 17:19
1
$begingroup$
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.
$endgroup$
– The Photon
Dec 22 '18 at 18:22
1
$begingroup$
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.
$endgroup$
– Philip Wood
Dec 22 '18 at 18:27
|
show 3 more comments
$begingroup$
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 '18 at 16:44
akhmeteliakhmeteli
17.7k21841
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "151"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f449856%2fnormal-force-not-perpendicular-to-the-surface%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
"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 '18 at 16:41
Philip WoodPhilip Wood
7,9733616
$endgroup$
1
$begingroup$
Contact force is the vector sum of the frictional force and the normal force so even that is incorrect terminology.
$endgroup$
– harshit54
Dec 22 '18 at 17:04
1
$begingroup$
@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!
$endgroup$
– Philip Wood
Dec 22 '18 at 17:16
1
$begingroup$
Oh, I just saw that they write the formula as $F_n=mg_{eff}$ so that takes the rotation into account. My apologies.
$endgroup$
– harshit54
Dec 22 '18 at 17:19
1
$begingroup$
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.
$endgroup$
– The Photon
Dec 22 '18 at 18:22
1
$begingroup$
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.
$endgroup$
– Philip Wood
Dec 22 '18 at 18:27
|
show 3 more comments
$begingroup$
"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 '18 at 16:41
Philip WoodPhilip Wood
7,9733616
$endgroup$
1
$begingroup$
Contact force is the vector sum of the frictional force and the normal force so even that is incorrect terminology.
$endgroup$
– harshit54
Dec 22 '18 at 17:04
1
$begingroup$
@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!
$endgroup$
– Philip Wood
Dec 22 '18 at 17:16
1
$begingroup$
Oh, I just saw that they write the formula as $F_n=mg_{eff}$ so that takes the rotation into account. My apologies.
$endgroup$
– harshit54
Dec 22 '18 at 17:19
1
$begingroup$
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.
$endgroup$
– The Photon
Dec 22 '18 at 18:22
1
$begingroup$
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.
$endgroup$
– Philip Wood
Dec 22 '18 at 18:27
|
show 3 more comments
$begingroup$
"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 '18 at 16:41
Philip WoodPhilip Wood
7,9733616
$endgroup$
"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 '18 at 16:41
Philip WoodPhilip Wood
7,9733616
edited Dec 23 '18 at 9:08
answered Dec 22 '18 at 16:41
Philip WoodPhilip Wood
7,9733616
answered Dec 22 '18 at 16:41
Philip WoodPhilip Wood
7,9733616
answered Dec 22 '18 at 16:41
Philip WoodPhilip Wood
7,9733616
7,9733616
1
$begingroup$
Contact force is the vector sum of the frictional force and the normal force so even that is incorrect terminology.
$endgroup$
– harshit54
Dec 22 '18 at 17:04
1
$begingroup$
@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!
$endgroup$
– Philip Wood
Dec 22 '18 at 17:16
1
$begingroup$
Oh, I just saw that they write the formula as $F_n=mg_{eff}$ so that takes the rotation into account. My apologies.
$endgroup$
– harshit54
Dec 22 '18 at 17:19
1
$begingroup$
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.
$endgroup$
– The Photon
Dec 22 '18 at 18:22
1
$begingroup$
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.
$endgroup$
– Philip Wood
Dec 22 '18 at 18:27
|
show 3 more comments
1
$begingroup$
Contact force is the vector sum of the frictional force and the normal force so even that is incorrect terminology.
$endgroup$
– harshit54
Dec 22 '18 at 17:04
1
$begingroup$
@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!
$endgroup$
– Philip Wood
Dec 22 '18 at 17:16
1
$begingroup$
Oh, I just saw that they write the formula as $F_n=mg_{eff}$ so that takes the rotation into account. My apologies.
$endgroup$
– harshit54
Dec 22 '18 at 17:19
1
$begingroup$
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.
$endgroup$
– The Photon
Dec 22 '18 at 18:22
1
$begingroup$
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.
$endgroup$
– Philip Wood
Dec 22 '18 at 18:27
1
1
$begingroup$
Contact force is the vector sum of the frictional force and the normal force so even that is incorrect terminology.
$endgroup$
– harshit54
Dec 22 '18 at 17:04
$begingroup$
Contact force is the vector sum of the frictional force and the normal force so even that is incorrect terminology.
$endgroup$
– harshit54
Dec 22 '18 at 17:04
1
1
$begingroup$
@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!
$endgroup$
– Philip Wood
Dec 22 '18 at 17:16
$begingroup$
@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!
$endgroup$
– Philip Wood
Dec 22 '18 at 17:16
1
1
$begingroup$
Oh, I just saw that they write the formula as $F_n=mg_{eff}$ so that takes the rotation into account. My apologies.
$endgroup$
– harshit54
Dec 22 '18 at 17:19
$begingroup$
Oh, I just saw that they write the formula as $F_n=mg_{eff}$ so that takes the rotation into account. My apologies.
$endgroup$
– harshit54
Dec 22 '18 at 17:19
1
1
$begingroup$
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.
$endgroup$
– The Photon
Dec 22 '18 at 18:22
$begingroup$
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.
$endgroup$
– The Photon
Dec 22 '18 at 18:22
1
1
$begingroup$
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.
$endgroup$
– Philip Wood
Dec 22 '18 at 18:27
$begingroup$
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.
$endgroup$
– Philip Wood
Dec 22 '18 at 18:27
|
show 3 more comments
$begingroup$
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 '18 at 16:44
akhmeteliakhmeteli
17.7k21841
$endgroup$
add a comment |
$begingroup$
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 '18 at 16:44
akhmeteliakhmeteli
17.7k21841
$endgroup$
add a comment |
$begingroup$
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 '18 at 16:44
akhmeteliakhmeteli
17.7k21841
$endgroup$
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 '18 at 16:44
akhmeteliakhmeteli
17.7k21841
answered Dec 22 '18 at 16:44
akhmeteliakhmeteli
17.7k21841
answered Dec 22 '18 at 16:44
akhmeteliakhmeteli
17.7k21841
answered Dec 22 '18 at 16:44
akhmeteliakhmeteli
17.7k21841
17.7k21841
add a comment |
add a comment |
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f449856%2fnormal-force-not-perpendicular-to-the-surface%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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.)
$endgroup$
– psitae
Dec 22 '18 at 18:37
$begingroup$
I vote you are correct and the problem is either misguided or badly worded.
$endgroup$
– ja72
Dec 22 '18 at 20:43