My teacher told me that there is no effects on the gravitational acceleration from the horizontal motion....

 

For eg, if we throw a ball horizonally in the same height as we drop the ball, both should reach the ground at the same time....

 

but I and my mate did an experiment. We put a trolley on a slope of a certain angle, although the trolley has constant acceleration, but it is alot different than the gravitational acceleration!!

 

Any body can explain the principle behind it??

 

Albert

The acceleration in the direction of the slope is not the same as the gravitational acceleration. it's going to be g* sin (inclination angle)

 

In the vertical direction, you no longer have just g as an acceleration, you also have the contribution from the vertical component of the normal force, which will be g*cos(inclination angle). So the vertical acceleration will end up being g(1-cos(inclination angle)).

Then how it differ to my teacher's explanation??

 

Albert

By the way, apart from my previous post, why the acceleration from gravity changes due to the inclination angle??

 

Albert

The horizontal motion doesn't affect g. The incline affects the vertical acceleration because it adds a force along that axis. vertical acceleration and g aren't the same thing. Neither are "horizontal motion" and "inclined plane."

so what is that vertical acceleration??? is'nt it due to gravity???

During my experiment, I just placed the trolley on a slope, and simply let loose my hand, it went...

 

Albert

sorry for being not understandable..., but this is my first time dealing with the change of acceleration due to the inclined angle....

 

This is an useful site: http://lectureonline.cl.msu.edu/~mmp/applist/si/plane.htm

 

but also where I need explanation from...

 

for the red arrow, there is no question, it is the force due to gravity....

 

but what I dont get are those two black arrows, why the force due to gravity change its direction for those?? I mean, the gravitational force is always downwards, if it can go in those directions, can the gravity force goes upwards direction????

 

such confuse.....

 

Any body can help??

 

Albert

What Swatson means is this:

 

If the slope is at an inclination of theta to the horizontal,

the vertical acceleration consists of two components.

 

One component is the acceleration due to gravity g which is acting downwards.

The other component is your normal reaction force acting on the trolley by the slope. This normal force is always acting at right angles on the trolley upwards, and it will always be the same magnitude as the force that is acting downwards perpedicular to the slope so that you will not see the trolley tunnelling into the slope. This normal force will then have the magnitude of

g*cos(theta), using the above reasoning. The normal force component in the vertical direction will then be g*cos^2(theta), resolving the component of the normal force in the vertical direction. So the vertical acceleration will have g in the downward direction, and g*cos^2(theta) in the upward vertical direction.

 

So the effective reduced vertical accleration will be

g - g*cos^2(theta) m/s^2 in the downward direction.

In brief, when you throw something with horizontal force applied, it moves away you and reach the ground.

On the other hand, you let an object falling vertically. The times required for them to reach the ground are the same ( neglect air resistance).

sorry for being not understandable...' date=' but this is my first time dealing with the change of acceleration due to the inclined angle....

 

This is an useful site: http://lectureonline.cl.msu.edu/~mmp/applist/si/plane.htm

 

but also where I need explanation from...

 

for the red arrow, there is no question, it is the force due to gravity....

 

but what I dont get are those two black arrows, why the force due to gravity change its direction for those?? I mean, the gravitational force is always downwards, if it can go in those directions, can the gravity force goes upwards direction????

 

such confuse.....

 

Any body can help??

 

Albert[/quote']

 

The force due to gravity is always down in those diagrams - it doesn't change. That's a red arrow - always down.

 

However, once the plane is inclined, the force on the object in the coordinate system of the plane, is no longer just Mg - it's ^{[math]Mg cos\theta[/math]} perpendicular to the plane, and ^{[math]Mg sin\theta[/math]} parallel to the plane. (if you add those vectorally you still get Mg as the total)

 

You need to get used to using different coordinate systems - it ultimately makes solving the problem easier.

I still dont get the principle.....

 

You guys tell me about how to find the force when the plane is inclined...

 

but I want to ask why....

 

Since gravitational force is first applied on the trolley downwards, the slope at whatever angle, should act with the same amount of force reversely (upwards)...

Why the direction is changed??? I never know that g's direction can be diagonal....

 

I know how to calculate, but I want to know why does this happen.... Maybe a picture with arrows, numbers, and caculations would dexcribe better??

 

thx in advance

 

Albert

You are mixing up the gravitational force and the net force.

 

The gravitational force never changes.

 

The net force, which dictates the acceleration of the object, is the vector sum of all the forces - gravitational and normal force, and any others that might be present.

is the force moving diagonally downwards the resultant force??

The resultant force is in the direction of the acceleration. In this case it's parallel to the incline.

If you still can't get it, you may draw a picture with details to ask more clearly.

That's better.

 

Above is the picture of what I think instead of the picture in the site....

 

the Fg is the action that first pull every particle of the object downwards....

The slope, having an action on it, gives a reaction back with equal amount of force upwards...

 

Any more explanation towards the topic??

 

Albert

the reaction force will always be perpendicular to the slope.

the reaction force will always be perpendicular to the slope.

 

No, the reaction force is always equal in magnitude and opposite in direction, by Newton's third law. But that's irrelevant - the reaction force is the gravitational force the block exerts on the earth, and we don't care about that.

 

The free-body diagram needs to show all the forces acting on the block. The surface exerts a normal force, which is (by definition) normal, or perpendicular, to the surface.

So, Swan, Why there is a force exerting perpendicularly from the slope to the object, and a reaction back to the slope??? I think there are no such forces.....

 

Can you tell me why??

 

Albert

oops sorry. got confused with my reaction forces and normal forces.

oops sorry. got confused with my reaction forces and normal forces.

So' date=' Swan, Why there is a force exerting perpendicularly from the slope to the object, and a reaction back to the slope??? I think there are no such forces.....

 

Can you tell me why??

 

Albert[/quote']

 

That's true of any contact between surfaces. You don't think your chair is pushing back up at you when you sit in it?

So' date=' Swan, Why there is a force exerting perpendicularly from the slope to the object, and a reaction back to the slope??? I think there are no such forces.....

 

Can you tell me why??

 

Albert[/quote']

 

That's true of any contact between surfaces. You don't think your chair is pushing back up at you when you sit in it?

The weight of an object is a force due to gravity towards the centre of the earth.

Normally, it is perpendicular to the ground. It is fundatmental and continuous force.

 

About the normal force....

When you stand up, you are not pulled by the strong fundatmenal gravitional force.

According to Newton First Law, there is no net force. Obviously, that force acts against the gravitational force is the force by the ground. It is a reaction force which maintain your equillibrium and it is passive, so it is called normal force. Actually, it is a kind of electromagnetic force.

Down a slope, there is a force from the surface of that pyramid. That's the normal force. Fancy without this force, the object will go into the pyramid since there is a net force.

The weight of an object is a force due to gravity towards the centre of the earth.

Normally, it is perpendicular to the ground. It is fundatmental and continuous force.

 

About the normal force....

When you stand up, you are not pulled by the strong fundatmenal gravitional force.

According to Newton First Law, there is no net force. Obviously, that force acts against the gravitational force is the force by the ground. It is a reaction force which maintain your equillibrium and it is passive, so it is called normal force. Actually, it is a kind of electromagnetic force.

Down a slope, there is a force from the surface of that pyramid. That's the normal force. Fancy without this force, the object will go into the pyramid since there is a net force.