Almost Physics I  Gravitational Force

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 Joined: Mon Nov 04, 2013 7:10 pm
My posts are covering basic physics laws, I hope readers are aware that I am still newbie and that what you read in my post is far away to replace the Marshals, Explorers and Rescue team teaching advises through the organized trips. I discourage readers to keep the content of this posts as a standard when driving as the desert may be an hazardous terrain, the beautiful straight dune that you might see on a drawing maybe completely different in real. So please, what you read is just there for general culture.
Gravitational Force
As it says, it is a force (F). This force has a vector, meaning that it has a direction. This direction is pointed toward the largest body. Here we compare two bodies, our vehicle and earth globe. So this vector is pointed towards earth core.
What affect the gravitational force in our case is the mass of our vehicle (m) and the gravity acceleration (g) = 9,81 m/s^{2}
The gravity acceleration 9.81m/s^{2} is also a vector pointed toward earth core and it means that if you launch your car jack in the air from the fossil rock, from the point where it start to fall, it will gain a speed of 9.81m/s each second. So after 2 second its speed will be 9.81*2 = 19.62m/s = 70.63 km/h. If we replace the car jack with our vehicle, this fall speed will remain the same. (In this example we don't consider the air friction force which normally reduces the fall speed depending how big the falling object is.) You can try this with an apple, not the car jack of course.
The relation between these values is F = m * g
Example: our vehicle mass is 2500 kg, the gravitational force will be 2500 * 9.81 = 24,525.00 Newton
This force is spread on the ground through the vehicle four tires.
If we imagine that 60% of our vehicle mass is concentrated on the front wheels because of the engine weight and the remaining 40% on the rear wheels. The calculation will be :
Front 24525 N * 0.60 / 2 tires = 7357.5 N gravitational force on each of the front tire.
Rear 24525 N * 0.40 / 2 tires = 4905 N gravitational force on each of the rear tire.
This is why when sometimes we look at our vehicle tires at flat road, the front tires looks a little bit deflated compared to the rear ones. Actually they are not deflated, but there is more force applied downward on the front tires than the rear ones.
Another example is when we make a car jump from the crest of a dune:
https://drive.google.com/file/d/0B8Nggl ... sp=sharing
On the image (link above) the car started to rotate on it self, no matter what the jump curve is, following the equation F = m * g, so the more mass we have, the first falling is. This is because of the force applied on the car at the front is higher than the one applied on the rear.
The gravitational force also helps to descend a dune but it is only one of many criteria that we will discuss in the next threads
I hope this helps to understand the basic of gravitational force, we will be discussing in the next thread the normal force which explain why our car is not falling into the ground.
Regards,
Mohanned
Any correction or clarification are most welcome.
Links: en.wikipedia.org/wiki/gravitational_acceleration
Gravitational Force
As it says, it is a force (F). This force has a vector, meaning that it has a direction. This direction is pointed toward the largest body. Here we compare two bodies, our vehicle and earth globe. So this vector is pointed towards earth core.
What affect the gravitational force in our case is the mass of our vehicle (m) and the gravity acceleration (g) = 9,81 m/s^{2}
The gravity acceleration 9.81m/s^{2} is also a vector pointed toward earth core and it means that if you launch your car jack in the air from the fossil rock, from the point where it start to fall, it will gain a speed of 9.81m/s each second. So after 2 second its speed will be 9.81*2 = 19.62m/s = 70.63 km/h. If we replace the car jack with our vehicle, this fall speed will remain the same. (In this example we don't consider the air friction force which normally reduces the fall speed depending how big the falling object is.) You can try this with an apple, not the car jack of course.
The relation between these values is F = m * g
Example: our vehicle mass is 2500 kg, the gravitational force will be 2500 * 9.81 = 24,525.00 Newton
This force is spread on the ground through the vehicle four tires.
If we imagine that 60% of our vehicle mass is concentrated on the front wheels because of the engine weight and the remaining 40% on the rear wheels. The calculation will be :
Front 24525 N * 0.60 / 2 tires = 7357.5 N gravitational force on each of the front tire.
Rear 24525 N * 0.40 / 2 tires = 4905 N gravitational force on each of the rear tire.
This is why when sometimes we look at our vehicle tires at flat road, the front tires looks a little bit deflated compared to the rear ones. Actually they are not deflated, but there is more force applied downward on the front tires than the rear ones.
Another example is when we make a car jump from the crest of a dune:
https://drive.google.com/file/d/0B8Nggl ... sp=sharing
On the image (link above) the car started to rotate on it self, no matter what the jump curve is, following the equation F = m * g, so the more mass we have, the first falling is. This is because of the force applied on the car at the front is higher than the one applied on the rear.
The gravitational force also helps to descend a dune but it is only one of many criteria that we will discuss in the next threads
I hope this helps to understand the basic of gravitational force, we will be discussing in the next thread the normal force which explain why our car is not falling into the ground.
Regards,
Mohanned
Any correction or clarification are most welcome.
Links: en.wikipedia.org/wiki/gravitational_acceleration