Calculating Time Needed for Box to Stop Sliding Down Slope

In summary, a box will stop before reaching the end of a slope when the coefficient of friction is 0.1 and the acceleration is 0.
  • #1
crom1
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Box is sliding down the slope with angle 30 , and v_0=0. Because coefficient of friction is $$\mu=0.1x$$ where x is distance covered, the box will stop before reaching the end of a slope. Find the time needed for a box to stop.

I get $$F_1=G \sin \alpha , F_{fr} = \mu \cdot G \cos \alpha$$ and the box will stop when $$F_1 = F_{fr}$$ , that is x=5.77. Ok, so I think I got the distance x, but I have no idea how to find time t. Acceleration is obviously changing, and because of that I'm not sure what formulas can I use, also I have acceleration a as linear function of x (not t) , I tried with some integrating but didn't lead me no where (I was probably going in circles there).
Any hint on how to find time t?
 
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  • #2
crom1 said:
, I tried with some integrating but didn't lead me no where (I was probably going in circles there).
That should be the way to go. Please post your working.
 
  • #3
crom1 said:
and the box will stop when ##F_1 = F_{fr}##
No, think about that again.
 
  • #4
If the part with F_1=F_{fr} is wrong, I don't think there's a point in posting my working before I realize when will the box stop. Hmm, can you show mistake in my reasoning?
Since v_0=0 , the box will stop when v=0 , and v depends on acceleration, so acceleration has to be 0 , so F{res}/m=0 , and F_1=F_{fr} .
 
  • #5
crom1 said:
If the part with F_1=F_{fr} is wrong, I don't think there's a point in posting my working before I realize when will the box stop. Hmm, can you show mistake in my reasoning?
Since v_0=0 , the box will stop when v=0 , and v depends on acceleration, so acceleration has to be 0 , so F{res}/m=0 , and F_1=F_{fr} .
Yes, acceleration will be zero when the two forces are equal, but what does that tell you about the velocity?
 
  • #6
velocity won't change, and acceleration has to be negative in order to v be 0 at some moment?
Okay, but how do I calculate velocity when acceleration is changing and I don't have velocity in terms of t ? I tried something like this:

$$ a=g \sin \alpha - \mu g \cos \alpha = g \sin \alpha-0.1 g \cos \alpha x = -cx+d $$
$$ \frac{dv}{dt} = -cx+d $$
$$ \frac{dv}{-cx+d}= dt \Rightarrow t= \frac{v}{-cx+d} $$ but doesn't work
 
  • #7
crom1 said:
$$ \frac{dv}{dt} = -cx+d $$
Right. You don't recognise that equation? What if you write that in terms of ##\ddot x##? You can get rid of the d by a simple change of variable.
 
  • #8
I'm currently learning this on my own, so I might miss some obvious and basic stuff. Do you mean like this

$$ \frac{d^2x}{dt^2}= -cx+d $$ (Did you meant some kind of substitution y=x-d/c or? )

EDIT: I did get t=-1/c ln|-cx+d| but I don't have x so that probably not good...
 
Last edited:
  • #9
crom1 said:
(Did you meant some kind of substitution y=x-d/c or? )
Yes, exactly that.
What do you know about SHM?
 
  • #10
Simple harmonic motion? Yes, I see that this is the same type of equation. We didn't yet talked about harmonic motion in my physics course (We did in high school but on that level) so don't be suprised by my ignorance :D , but one more question : when I get x in terms of t, how should I find t when I don't know x?
Or how do I find x?
 
  • #11
crom1 said:
Simple harmonic motion? Yes, I see that this is the same type of equation. We didn't yet talked about harmonic motion in my physics course (We did in high school but on that level) so don't be suprised by my ignorance :D , but one more question : when I get x in terms of t, how should I find t when I don't know x?
Or how do I find x?
You can find x without invoking SHM. You only need SHM for the time. Consider work.
Or, using the SHM solution, you can use the initial velocity as the boundary condition.
 

Related to Calculating Time Needed for Box to Stop Sliding Down Slope

1. How do you calculate the time needed for a box to stop sliding down a slope?

The time needed for a box to stop sliding down a slope can be calculated using the formula t = √(2d/gμ) where t is the time, d is the distance traveled, g is the acceleration due to gravity, and μ is the coefficient of friction.

2. What is the coefficient of friction and how does it affect the time needed for the box to stop?

The coefficient of friction is a unitless value that represents the resistance between two surfaces in contact. It affects the time needed for the box to stop as a higher coefficient of friction means there is more resistance, resulting in a longer time to stop.

3. Can this formula be applied to any slope angle?

This formula can be applied to any slope angle as long as the angle does not exceed the angle of repose, which is the steepest angle at which a box can rest without sliding down the slope.

4. Is this formula only applicable to boxes?

No, this formula can be applied to any object sliding down a slope as long as the object's weight, distance traveled, and coefficient of friction are known.

5. Are there any other factors that can affect the time needed for the box to stop sliding?

Other factors that can affect the time needed for the box to stop sliding include the surface material of the box and the slope, air resistance, and the shape and weight distribution of the box.

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