Hanging box - radius, length, mass, speed

AI Thread Summary
A lunch box suspended in a school bus hangs at an angle of 28 degrees with the vertical while the bus travels in a circular path with a radius of 48.16 m. The problem involves calculating the radius of the circular path the lunch box makes, initially leading to confusion about the correct application of trigonometric functions. After several calculations, the radius of the circle made by the lunch box was determined to be approximately 1.57 m. Participants also discussed how to find the tension in the string, emphasizing the need to analyze forces using Newton's laws. The discussion concluded with the participant successfully determining the tension after clarifying the forces acting on the box.
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Hanging box -- radius, length, mass, speed

Homework Statement



You are riding in a school bus. As the bus rounds a flat curve at constant speed, a lunch box with a mass of 0.490 kg suspended from the ceiling of the bus by a string of length 1.79 m is found to hang at rest relative to the bus when the string makes an angle of 28.0^\circ with the vertical. The point (on the ceiling of the school bus) to which the string is tied is traveling in a circle of radius 48.16 m.

The lunch box is also traveling in a circular path; what is the radius of the circle in which the lunch box is traveling?

Homework Equations



n/a

The Attempt at a Solution



I drew a little sketch.

a horizontal line (bus roof)

a vertical line starting in the middle of the horz line that goes down 1.79 m

then i drew a line at bottom of the vertical line to the right and labeled that (x)

i made a hypotenuse line that goes from the starting point on the vertical line to the end on the x line.

This made a right triangle. If the degree between the horizontal and the string is 28 degrees then the angle that is down by the 90 degree angle should be 28 degrees as well correct?

so to solve for x i did

Tan(28) = 1.79/x

x = 3.36m

i believe this will be the radius of the circle it is making but I am not sure.

Can anyone let me know if this is correct please :)

Thank you
 
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If the string is 1.78m long then it should be the hypotenuse which is also 1.78m long, not the vertical line.
 


correct...

so is it more like

1.78cos(28) = x

x= 1.57 m? is that the radius?
 


Don't forget: the string makes an angle of 28.0^\circ with the vertical. In your work you decided to change that so it's 28 degrees to the horizontal for some reason. (I'm hoping I read the question correctly, this is my first time on the forum)
 


ahhh thank you i didnt even notice that. yeah it is with the vertical but i think i got it now.

I got 1.79*cos(62) = x

x = 0.840 m

that seems a little small does that look right for the radius of the circle that the lunch box is making?
 


It looks right to me!
 


Thank you.

Can you maybe help me out with the next part? it deals with tension and i think you have to involve the centrip. accel of the bus and I am just not sure what to do.

What is the tension in the string?
 


my online homework just said 0.84 m was wrong? hmmmm...
 


mybrohshi5 said:
my online homework just said 0.84 m was wrong? hmmmm...
What happened to the 48.16 m radius of the 'point on the ceiling' circle ?
 
  • #10


what do you mean what happened to it? Do i have to include that radius that the bus is traveling into figure out the radius that the string is making?
 
  • #11


I believe so
 
  • #12


that doesn't make any sense to me? so is the question asking me to add the radius i just found to the radius of the circle that the bus is making?

AHHHH i hate mastering physics :(
 
  • #13


Can anyone help me with this one? its my last one on this homework assignment and i can't figure it out :(

I need to find the radius that the string makes first and then i need to find the tension of the string and then a few other parts...

please help if anyone knows about this type of problem :)

thank you
 
  • #14


So surely by now you know the radius. The tension in the string can be found by drawing a free body diagram of the box. Identify the forces, known and unknown, acting on it. Then apply Newton's laws to solve for T. When you apply Newton's law in the vertical direction, is the box accelerating in the vertical direction?
 
  • #15


Yes i found the radius :)

So for tension there won't be any acceleration in the vertical direction only in the horizontal...
 
Last edited:
  • #16


never mind i got the tension :)

thanks :)
 
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