# Force question

allstar1
Is the normal force on a body always equal to its weight?

YES or NO?

I believe the answer is NO.

Gold Member
You are correct.

allstar1
Can you give me your reason for it??

I believe the answer is NO, because the normal force on a body is only equal to its weight when the body is on a horizontal surface.

et me know if my reasoning is correct. Thanks.

Dorothy Weglend
allstar1 said:
I believe the answer is NO, because the normal force on a body is only equal to its weight when the body is on a horizontal surface.

et me know if my reasoning is correct. Thanks.

If you have a body on an incline, then the force of gravity on that body would be F_grav = mg (cos a i + sin a j). But if you compute the magnitude of this vector, you end up with |F_grav| = mg, don't you?

Doesn't this mean that the weight of a body on an incline is the same as on a horizontal surface?

Dot

Mentor
allstar1 said:
I believe the answer is NO, because the normal force on a body is only equal to its weight when the body is on a horizontal surface.
If you are talking about an object just resting on a surface, then you are correct.

Mentor
Dorothy Weglend said:
If you have a body on an incline, then the force of gravity on that body would be F_grav = mg (cos a i + sin a j). But if you compute the magnitude of this vector, you end up with |F_grav| = mg, don't you?
Putting a body on an incline doesn't change the direction or magnitude of its weight. It's always just $mg$ acting down.

Doesn't this mean that the weight of a body on an incline is the same as on a horizontal surface?
It better be the same!

Dorothy Weglend
Doc Al said:
Putting a body on an incline doesn't change the direction or magnitude of its weight. It's always just $mg$ acting down.

It better be the same!

Ah, right. I understand. I misunderstood the original question. Yes, the normal force is less on an incline.

I've been thinking of the inclines in these mechanics problems as splitting the downard force of mg into two components, one parallel to the incline, and one perpendicular to it. So I guess I have been thinking of the incline as a device to "split gravity", although I know that sounds stupid. I haven't really used that term until now.

Could you enlarge on what is really going on? I'd appreciate it.

Dot

Mentor
For an object on an incline it is often useful to find the components of the weight parallel and perpendicular to the surface. If the incline is fixed, only forces parallel to the incline will accelerate the object. (The perpendicular component will equal the normal force.)

allstar1
the question states "Always equall", i guess the always is what makes the difference. Because the Normal force is only equal when object is at rest horizontally. I guess if it be on an incline then ther object would slide down and move. Therefore on object inclined would NOT have an equal normal force.

That's just my opinion.

Dorothy Weglend
Doc Al said:
For an object on an incline it is often useful to find the components of the weight parallel and perpendicular to the surface. If the incline is fixed, only forces parallel to the incline will accelerate the object. (The perpendicular component will equal the normal force.)

I understand this. What I don't understand is why this isn't viewed as changing the direction of the objects weight. I've been told I get too hung up on the details, but I do like to understand these things.

Dot

Mentor
Finding the components of a force does not change the force. Weight is defined as the gravitational force that the Earth exerts on an object. That doesn't change just because an object is on an incline. (Of course, being on an incline does change the forces on the object, but not its weight.)