Block Acceleration up an Inclined Plane

  • #1

Homework Statement



A cable is pulling a 200kg block up an inclined plane elevated at 30 degrees with 3.5KN of force. How fast is the block accelerating at?

θ = 30
g = 9.81
m = 200kg
μ = 0.35
F = kN

Homework Equations



F = ma

The Attempt at a Solution



Wasn't sure how to do this problem, I somehow got this...

ma = m*g*sinθ - μ*m*g*cosθ

But don't know how to implement the 3.5kN force into it.
 
Last edited:

Answers and Replies

  • #2
881
40

Homework Statement



A cable is pulling a 200kg block up an inclined plane elevated at 30 degrees with 3.5KN of force. How fast is the block accelerating at?

θ = 30
g = 9.81
m = 200kg
μ = 0.35
F = kN

Homework Equations



F = ma

The Attempt at a Solution



Wasn't sure how to do this problem, I somehow got this...

ma = m*g*sinθ - μ*m*g*cosθ

But don't know how to implement the 3.5kN force into it.

What you have written is the force equation. It gives the sum of all forces on the body, in direction of the incline, to be equal to the acceleration(in that direction) times its mass, from Newton's second law. Do you see where your 3.5kN force would join in now?
 
  • #3
28
0
Remember that forces are vectors and can act in different directions. Think about which forces act in which direction (Up or down the hill) in the problem. The 3.5kN work to accelerate the block .... the hill, gravity and friction work to accelerate the block .... the hill. Together they make a net force in a certain direction.
 
  • #4
1,065
10
Good approach is to draw fbd.
 
  • #5
so basically, what I would do is use that equation I've got down there:

ma = m*g*sinθ - μ*m*g*cosθ

200a = 200*9.81*sin(30) - 0.35*200*9.81*cos(30)

a = 1.9315 down hill

Then... F = ma... = 200 * 1.9315 = 386.3

3500 - 386.3 = 200*a(uphill)

a = 15.5685m/s^2

Which is ridiculously fast, but, is a fair call because it's only a 200kg mass with a 3.5 tonne force.

So 15.5685m/s^2...?
 
  • #6
28
0
Take azizlwl's advice and draw a force diagram.

That being, if you don't know, a simplified picture of the situation showing planes, masses and forces and any relevant angles and figures. In this case you would draw a box sitting on an inclined plane, fill in the angle of the plane and any known figures (The mass of the block, for example)

Having drawn this, think about all the forces in the situation and in which direction they act. Clearly you understand that gravity acts downwards and a component of this equal to mgsinθ acts down the hill. You also clearly understand that the resistance force as a result of friction is equal to mg0.35cosθ.

You've given your equation as ma = m*g*sinθ - μ*m*g*cosθ

Questions; Which direction does the resisting frictional force act? Is it the same or different direction to the weight force? Do forces acting in the same direction combine with or cancel each other?
In which direction does the 3.5kN force act? Is it the same or difference to the weight and frictional forces?
How can you mathematically represent the different directions in which a force acts?

^ Hopefully that jogs your mind along a little. Additionally, I'd avoid referring to force in terms of mass "3.5 tonne force".
 
  • #7
1,065
10
so basically, what I would do is use that equation I've got down there:

ma = m*g*sinθ - μ*m*g*cosθ

200a = 200*9.81*sin(30) - 0.35*200*9.81*cos(30)

a = 1.9315 down hill

Then... F = ma... = 200 * 1.9315 = 386.3

3500 - 386.3 = 200*a(uphill)

a = 15.5685m/s^2

Which is ridiculously fast, but, is a fair call because it's only a 200kg mass with a 3.5 tonne force.

So 15.5685m/s^2...?

ma = m*g*sinθ - μ*m*g*cosθ
Your frictional force is going uphill.
3500 - 386.3 = 200*a(uphill)
Which is direction of the motion.
 
  • #8
28
0
ma = m*g*sinθ - μ*m*g*cosθ
Your frictional force is going uphill.
3500 - 386.3 = 200*a(uphill)
Which is direction of the motion.

Don't make it too easy..
 
  • #9
So instead of

ma = m*g*sinθ - μ*m*g*cosθ

it should be...

ma = m*g*sinθ + μ*m*g*cosθ

Which'll give me 1925N of force downwards...

Then doing the same thing, I'll have

3500 - 1925 = 200a
a = 9.625m/s^2
 
  • #11
Ugh all because of one mistake which I thought wasn't D: Thank you so much!
 
  • #12
Could I use the same principles for this question? http://img844.imageshack.us/img844/6641/f5fec6746b9c42289610ba1.png [Broken]
 
Last edited by a moderator:
  • #13
28
0
Yes, however don't expect any help from me on that one. I could sit down for a while and try it but I'm by no means a physics guru like some of the people here. While you can apply the same principles, there's a few other principles you need to apply. Like force to overcome static friction etc. And the whole thing is made slightly more complicated by the lack of workable figures.
 
  • #14
Mm, because, I have no idea really on how to implement static as well as kinetic friction into this problem :/
 
  • #15
28
0
Well static friction is the frictional force between the mass and the plane when the mass is stationary. For example if I had a 10kg mass and a Us of 0.1, the frictional force would be mg0.1. We'll take that to equal 10N because I'm feeling lazy and 9.81 is just too much.

So the frictional force is 10N, and that acts in resistance to the pulling force. Assuming The weight force down the incline is equal to say, 30N, we would need to apply a force of >40N to get the object moving.

Then, however, Us becomes invalid. We have to apply Uk. If Uk was equal to 0.05, making the resisting frictional force 5N only, the force required to accelerate the mass is >35N. Applying a force equal to 35N would merely keep the mass moving at its velocity at that instant.

Make sense?
 
  • #16
Should I actually be getting a figure here as an answer or an equation?
 
  • #17
28
0
Equation.
 

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