Physics help

1. Oct 9, 2004

Winner

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.

2. Oct 9, 2004

arildno

Welcome to PF!
What force must provide the centripetal acceleration of the block?

3. Oct 9, 2004

Winner

ok, hmm, force of gravity? lol wild guess here.

4. Oct 9, 2004

Winner

no wait, centripetal force then?

5. Oct 9, 2004

arildno

The centripetal acceleration is in the horizontal plane.
(Agreed?)
Hence, how much can the force of gravity contribute to the centripetal acceleration?

6. Oct 9, 2004

arildno

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. Oct 9, 2004

Winner

Yes, agreed to the first part. Gravity works vertically, so 0?

8. Oct 9, 2004

arildno

Correct!
So, what other force than gravity works on the block?

9. Oct 9, 2004

Winner

Maybe tension, or the normal force lol. This is great, it's like an online test. But still confused how this all fits.

10. Oct 9, 2004

arildno

What normal force????????????????????+
What should be the agent of that force?

So, state those two forces you believe acts on the block!

11. Oct 9, 2004

Winner

Wait, there's no normal force, just tension. This feels like punishment :| . So...how do I find theta now lol.

12. Oct 9, 2004

arildno

Misconceptions of physics ought to be whipped out of your mind, yes?
You're right, apart from gravity, we have the rope tension.
1.Now, suppose the rope makes an angle $$\theta$$ 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. Oct 9, 2004

arildno

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. Oct 9, 2004

Winner

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. Oct 9, 2004

arildno

But gravity is proportional to mass, right?
Go ahead with "m", and formulate Newton's 2.law in both directions..

16. Oct 9, 2004

Winner

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. Oct 9, 2004

arildno

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. Oct 9, 2004

Winner

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. Oct 9, 2004

arildno

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. Oct 9, 2004

Winner

ok, so..the sum of forces F=T-Fg?