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Is the normal force on a body always equal to its weight?
YES or NO?
I believe the answer is NO.
YES or NO?
I believe the answer is NO.
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?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 are talking about an object just resting on a surface, then you are correct.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.
Putting a body on an incline doesn't change the direction or magnitude of its weight. It's always just [itex]mg[/itex] acting down.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?
It better be the same!Doesn't this mean that the weight of a body on an incline is the same as on a horizontal surface?
Ah, right. I understand. I misunderstood the original question. Yes, the normal force is less on an incline.Doc Al said:Putting a body on an incline doesn't change the direction or magnitude of its weight. It's always just [itex]mg[/itex] acting down.
It better be the same!
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.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.)