Distance Travled in the x direction on a circular path given force

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 2K views
JarrodBrandt
Messages
2
Reaction score
0

Homework Statement


There is a block suspended by a string with length 6m forming the radius of a circle. The block is 220kg and you are able to exert a 420N force in the x direction. How far from the vertical will the block travel.

m=220kg
r=6m
F(x)=420

Find x(distance)


Homework Equations


F=ma
?

The Attempt at a Solution



What I attempted to do with this situation was to first find the vertical component of force, or 220*g which gives 2156N.
Setting the forces up to form a triangle, I found the hypotenuse to be 2196.5 by sqrt((420^2)+(2156^2)).
From here I found the angle from the vertical, or the top angle by tan-1(420/2156)=11.02 deg.
Next I applied this angle to the triangle dealing with the distance, taking sin(11.02)*6m=1.14m. So I have an answer, but I am unsure of the process I used to get it and if it is correct. Mostly I am unsure if the angle obtained from the forces "triangle" is applicable to the distance triangle. If I am correct, I'm not entirely sure why. Can someone help me with the theory involved here?
 
Physics news on Phys.org
What forces are present here?
 
voko said:
What forces are present here?

From what I can tell there is a force in the positive x direction of 420N, a force due to gravity of 2156N, and a normal force towards the center of the circle. My thought process initially was that to find the maximum x distance I would have to find the point where the forces add up to 0. I'm starting to think maybe I find this distance 1.14 then from there find the velocity of the object when it reaches that distance to find how much farther it will travel before gravity slows and eventually reverses it's direction. Anyone know if I'm in the right ballpark? or even playing the right game?
 
The normal force toward the center is the tension in the string. It must balance the projections of the force of gravity and of the motive force onto the string. Then the projections of two latter forces perpendicular to the string must balance each other.