What force must provide the centripetal acceleration of the block?

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  • #1
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physics help!!!

Ok here's the question:

A block is hung by a string from the inside roof of a van. When the van goes straight ahead at a speed of 28m/s, the block hangs verticaly down. But when the van maintains this same speed around an unbanked curve of radius 150m, the block swings toward the outside of the curve. Then the string makes an angle( theta) with the vertical. Find (theta)

Any suggestions? I can't seem to get anything out of this. The radii's are alwaays 150, pythagoras doesn't work. THe velocities are uniform, again pythagoras can't work. Maybe centripetal force? I can't use the banked curve formula, it's not banked!

Thanks,

Brian.
 

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  • #2
arildno
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Welcome to PF!
What force must provide the centripetal acceleration of the block?
 
  • #3
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ok, hmm, force of gravity? lol wild guess here.
 
  • #4
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no wait, centripetal force then?
 
  • #5
arildno
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The centripetal acceleration is in the horizontal plane.
(Agreed?)
Hence, how much can the force of gravity contribute to the centripetal acceleration?
 
  • #6
arildno
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Winner said:
no wait, centripetal force then?
We choose to call those forces (whatever else they are) which create centripetal acceleration for centripetal forces.
However, what is the basic force acting on the block which "plays the role" of centripetal force in this case?
 
  • #7
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Yes, agreed to the first part. Gravity works vertically, so 0?
 
  • #8
arildno
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Winner said:
Yes, agreed to the first part. Gravity works vertically, so 0?
Correct!
So, what other force than gravity works on the block?
 
  • #9
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Maybe tension, or the normal force lol. This is great, it's like an online test. But still confused how this all fits.
 
  • #10
arildno
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What normal force????????????????????+
What should be the agent of that force?

So, state those two forces you believe acts on the block!
 
  • #11
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Wait, there's no normal force, just tension. This feels like punishment :| . So...how do I find theta now lol.
 
  • #12
arildno
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Misconceptions of physics ought to be whipped out of your mind, yes? :wink:
You're right, apart from gravity, we have the rope tension.
1.Now, suppose the rope makes an angle [tex]\theta[/tex] to the vertical
2. What is the direction of the tensile force, and if the magnitude of the tensile force is "T", what is
a) The vertical component of the tensile force?
and
b) The horizontal component of the tensile force?
 
  • #13
arildno
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Assuming you get this, how must the VERTICAL component of Newton's 2.law look like?
And, what must the HORIZONTAL component of Newton's 2.law look like?
You've got two unknowns, the angle and the magnitude of the tension.
The two mentioned equations can be used to solve for the angle&magnitude of the tension force.
 
  • #14
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Something like Tsin(theta) = x comp, Tcos(theta)= y comp? I'm guessing tension acts upwards.

F=ma, but I don't have the mass.
 
  • #15
arildno
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But gravity is proportional to mass, right?
Go ahead with "m", and formulate Newton's 2.law in both directions..
 
  • #16
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ok, here's my guess. maSin(theta)=mv^2/r, that's for X and Y is just maCos(theta)? err, this is taking a while lol.
 
  • #17
arildno
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Does the block experience ANY SORT OF ACCELERATION IN THE VERTICAL DIRECTION, once it has reached the equilibrium state?

And yes, it DOES take a while to remove misconceptions..
 
  • #18
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ok, since you put it that way, no acc then. So it would just be Ty=Tcos(theta)? ahh, what's the answer? lol
 
  • #19
arildno
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Ok, now that you know that the acceleration is zero in the vertical, formulate Newton's 2.law for the balance of vertical forces..
 
  • #20
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ok, so..the sum of forces F=T-Fg?
 
  • #21
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Ok..I've spent way too much time for OnE question. Thanks for the help, I'll think about it.
 
  • #22
arildno
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I've outlined the procedure; you should be able to figure it out.
Besides, if you think the time necessary to remove your own misconceptions of physics is too much, well that's your problem.
 
  • #23
arildno
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For your information:
Vertical component of Newton's 2.law:
[tex]T\cos\theta-mg=0[/tex]
Horizontal component of Newton's 2.law
[tex]T\sin\theta=m\frac{v^{2}}{R}[/tex]
 
  • #24
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hmm, I was getting pretty close lol. Thanks again.
 
  • #25
Pyrrhus
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By the way

[tex]T\cos\theta = mg[/tex]

[tex]T\sin\theta=m\frac{v^{2}}{R}[/tex]

Divide to solve

[tex] \tan\theta = \frac{v^{2}}{Rg} [/tex]
 

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