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force, acceleration, and friction help1

 
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Nov21-11, 08:21 PM   #1
 

force, acceleration, and friction help1


1. The problem statement, all variables and given/known data

A space shuttle weighing 7.3 x 10^8 kg takes off from Cape Canaveral. If the air resistance is 1.2 x 10^5 N and the thrust is 8.9 x 10^11 N, how long will it take the shuttle to reach a velocity of 4500 m/s?


2. Relevant equations

at + vi = vf
ƩFy=mg
ƩFx=ma

3. The attempt at a solution

I know I have to subtract Fg and Ff (or are they the same thing?) from Fa (or N? normal force?) and set that equal to mass x gravity. Is gravity just -9.8m/s/s? And then I have to put that into at + vi = vf (acc. x time + initialvelocity = finalvelocity) but I'm just really lost.
 
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Nov21-11, 09:55 PM   #2
 
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Welcome to PF catandphysics!

Quote by catandphysics View Post
1. The problem statement, all variables and given/known data

A space shuttle weighing 7.3 x 10^8 kg takes off from Cape Canaveral. If the air resistance is 1.2 x 10^5 N and the thrust is 8.9 x 10^11 N, how long will it take the shuttle to reach a velocity of 4500 m/s?


2. Relevant equations

at + vi = vf
ƩFy=mg
ƩFx=ma
Careful. This bit in red is not correct (at least, not in general).

The thing that ALWAYS applies is Newton's 2nd law:

Fnet = ƩF = ma

(where I've used boldface to indicate vectors). This is the mathematical way of expressing the statement that "the net force acting on a body (i.e. the vector sum of all forces acting on it) is equal to the body's mass times its acceleration." We can apply this law separately for the horizontal and vertical directions:

ƩFx = max
ƩFy = may

We'll assume here that the shuttle's motion is entirely vertical, so that we can disregard the horizontal direction (i.e. this is essentially a 1D problem). So we just have the one equation for the vertical direction:

ƩF = ma

where I've dropped the 'y' subscripts since we're only dealing with one direction anyway. I've also dropped the vector notation, because in 1D it is sufficient represent the forces and the acceleration using signed scalars, where the sign tells you the direction (up or down).

So now we need to figure out what all of the vertical forces acting on the shuttle are, so that we can sum them up (and hence solve for the shuttle's acceleration 'a', which is what we are looking for). Draw a free body diagram of the shuttle, with all of the vertical forces acting on it represented. This is the best way to take an inventory of all the forces acting on the body -- it ensures that you don't miss anything.


Quote by catandphysics View Post
3. The attempt at a solution

I know I have to subtract Fg and Ff (or are they the same thing?) from Fa (or N? normal force?) and set that equal to mass x gravity. Is gravity just -9.8m/s/s? And then I have to put that into at + vi = vf (acc. x time + initialvelocity = finalvelocity) but I'm just really lost.
You are confused aren't you?

Ff usually refers to the frictional force, which is not relevant here, because the shuttle is not in contact with/sliding against another solid surface. On the other hand, you could think of air resistance (drag) as a "frictional force", so if you want to call it that, then that's fine. Fg is the gravitational force acting on the shuttle (i.e. it is the shuttle's weight). There is no normal force here, because the shuttle is in flight, rather than resting on the ground. The acceleration due to gravity is g = -9.81 m/s2.

Again, just apply Newton's 2nd law as I explained above. Take an inventory of all the vertical forces acting on the shuttle (from your FBD) and add them together (taking proper account of each of their signs, which tell you whether they point up or down). Set the result equal to ma. Solve for a.
 
Nov21-11, 10:05 PM   #3
 
Okay, so I do Fa - (Fg + Ff) = ƩF = ma, solve, and just plug that into vf^2 = vi^2 + 2aΔd. okay thanks so much for your help!
 
Nov21-11, 10:21 PM   #4
 
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force, acceleration, and friction help1


Quote by catandphysics View Post
Okay, so I do Fa - (Fg + Ff) = ƩF = ma, solve, and just plug that into vf^2 = vi^2 + 2aΔd. okay thanks so much for your help!
If Fa is the thrust, then I agree with that sum of forces. I'm assuming you used a subscript 'a' because it is the "applied" force from the engine.

You were right the first time. Once you know a, you want to plug it into vf = vi + at in order to solve for t. The other kinematic equation that you stated above won't help you, because it has no t in it, and t is the thing you are trying to find out.
 
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acceleration, forces, friction force, gravity, newton
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