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Sliding forces

by KMjuniormint5
Tags: forces, sliding
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KMjuniormint5
#1
Oct13-07, 11:04 PM
P: 67
The question is:

Two horizontal forces F1 and F2 act on a 1.8 kg disk that slides over frictionless ice, on which an xy coordinate system is laid out. Force F1 is in the negative direction of the x axis and has a magnitude of 3.0 N. Force F2 has a magnitude of 9.0 N. The figure below gives the x component vx of the velocity of the disk as a function of time t during the sliding. What is the angle between the constant directions of forces F1 and F2?

The graph is a Vs(m/s) vs t(s). It starts down at -4 m/s when time = 0 and it is a staight line up to +5m/s at time = 3 and the function crossed the x-axis around 1.3333ish (doesnt give an exact point).

how do I go about interupting this graph to help give me F2net . . .I think i know how to get the angle and I know that . . .

F1x = 3.0 N and F1y = 0 N so. . how to get F2x and F2y from the graph. .
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nnyms
#2
Jan13-08, 07:35 PM
P: 3
Let's list all given data.
- 1.8 kg disk
- F1 = -3.0 N (because it go in negative direction)
- F2 = 9.0 N (but we don't know which direction it go)
- a = ??? (it will be slope of your graph because your graph show velocity and if you derivative the velocity then you will get acceleration)

Use Newton's Second Law, Force = Mass x Acceleration.
The Force will be sum of F1 and F2. The Mass will be mass of disk. The Acceleration will be the slope of your velocity graph.

You will need to find F2x. (Note: the graph will not help you to find it.) Use cosine. The answer for F2x will be F2x = 9.0cos(d)

Set up an equation.
F1 + F2 = M x A
-3.0N + 9.0cos(d)N = 1.8kg x A

Go ahead find acceleration, then solve for angle 'd'. Let me know if you need any more hint and help.


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