MATLAB Answers

How to plot the velocity of a crate on a ramp for different angles?

7 views (last 30 days)
Kiera Harding
Kiera Harding on 8 Dec 2019
Edited: David Goodmanson on 9 Dec 2019
I have a task with a 12kg crate on a conveyor belt moving at u = 2.5m/s onto a ramp with kinetic friction uk = 0.3. The ramp is 3m long and I need to plot velocity against a range of angles then find the smallest angle for the ramp where the crate will still fall off at the end (s = 3).
I've been using angles from 0 to 30: A = linspace(0,30,301)
Then using this to get the normal force which seems to give correct values: N = m*g*cosd(A)
From this calculate the acceleration which also seems to give correct values: 12a = mgsinA - uk N -> a = (m*g*sind(A)-uk*N)/12
Then find the velocity using: v^2 = u^2 + 2as -> v = sqrt(u^2 + 2*a*s)
This gives imaginary values (e.g. 0.000 + 2.8267i) which I'm guessing due to this value being negative and then square root as the crate would realistically stop before reaching 3m. The values from 12.8 deg seem correct and don't have any imaginary parts.
When plotting A against v the velocity is 0 (as imaginary parts are ignored) until 10.8 when it starts to increase. At 10.8 deg the v value is 0.0000 + 0.2568i and at 10.9 deg it's 0.2018 + 0.0000i. When I've calculated it manually I get 22.6 degrees so unsure where I'm going wrong.
Any help is much appreciated!

  3 Comments

dpb
dpb on 8 Dec 2019
Makes no sense to use anything past the point at which the crate has moved past the end of the ramp...considering the center of mass to be the midpoint of the crate and at the endpoint as the tipping point would be reasonable approximation.
Image Analyst
Image Analyst on 8 Dec 2019
"How to plot the velocity" <== did you try plot():
plot(A, v, 'b-', 'LineWidth', 2); % Plot velocity vs. Angle
grid on;
Kiera Harding
Kiera Harding on 8 Dec 2019
I think this is correct - thank you for your help! I've calculated a few points seperately using the angles and I seem to be getting the right velocity for each one. I think my original calculation was completely off.

Sign in to comment.

Answers (1)

David Goodmanson
David Goodmanson on 9 Dec 2019
Edited: David Goodmanson on 9 Dec 2019
Hi Kiera,
The plot above shows the velocity of the box at the top of the ramp getting larger as the angle of the ramp gets steeper. Does this make sense physically? No. One problem is a sign error in the force. If displacement up the ramp is positive, the gravitational force m*g*sind(theta) is negative. The frictional force can be positive if the box is moving slower than the belt, so that the frictional force is helping to pull the box up hill. The equation is
v^2 = v0^2 + 2*a*s
with
a = (F/m) = (-m*g*sind(theta) + mk*N)/m
which is the same as your equation only with the force reversed. (Another way to express the acceleration is
a = g*(-sind(theta) + mk*cosd(theta))
).
The resulting plot shows the velocity at the top of the ramp getting smaller as the angle gets steeper. That's good, but like the first plot, this one also shows velocities greater than v0. The reason has to do with the mk*N term. It's in the equation all the time, but it really only applies when the box is sliding with respect to the belt, and that only happens when the gravitational force (the sine component, parallel to the belt) is greater than the available force of sliding friction. For small ramp angles the gravitational force is too small. The net result is that you can neglect any predicted velocity greater than v0, because those are produced by the mk*N term when the box is actually not sliding.

  0 Comments

Sign in to comment.

Sign in to answer this question.


Translated by