Calculating Acceleration for a Warehouse Ramp Pulley System with Friction

  • Thread starter Thread starter Mirole
  • Start date Start date
  • Tags Tags
    Pulley Ramp
AI Thread Summary
The discussion focuses on calculating the acceleration of a pulley system designed to move packages up a 15º ramp, with a counterweight twice the mass of the package. The user attempts to derive the acceleration using net force equations but struggles with incorporating the coefficient of kinetic friction, which is determined from a separate test where a 50 kg package is pushed with 250 Newtons. The calculated coefficient of friction is found to be approximately 0.51. Confusion arises regarding the correct application of mass in the equations, particularly whether to use the mass of the package or the combined mass of the system. The conversation emphasizes the need to set up two equations for the two masses involved to accurately solve for acceleration.
Mirole
Messages
33
Reaction score
0

Homework Statement


You have taken a summer job at a warehouse and have designed a method to help get heavy packages up a 15º ramp. In your system a package is attached to a rope which runs parallel to the ramp and over a pulley at the top of the ramp. After passing over the pulley the other end of the rope is attached to a counterweight which hangs straight down. In your design the mass of the counterweight is always adjusted to be twice the mass of the package. Your boss is worried about this pulley system. In particular, she is concerned that the package will be too difficult to handle at the top of the ramp and tells you to calculate its acceleration. To determine the influence of friction between the ramp and the package you run some tests. You find that you can push a 50 kg package with a horizontal force of 250 Newtons at a constant speed along a level floor made of the same material as the ramp.

Homework Equations


Fnet=ma


The Attempt at a Solution


Block 1:
(Fnet)y = n + T - Fgcos\theta= 0
(Fnet)y = n = mgcos\theta

(Fnet)x = n + T - Fgsin\theta - fk = ma
(Fnet)x = T - Fgsin\theta - fk = ma

(Fnet)x = 980 - Fgsin\theta - mu_k(mgcos\theta) = ma
(Fnet)x = 980 - gsinmgcos\theta - (mu_k)cos\theta = a

Block 2:
(Fnet)x = 0
(Fnet)y = T - 2mg = 0
(Fnet)y = T = 2*mg
(Fnet)y = T = 2*50*9.8, so T= 980N

I'm confused as to what to do next, because I don't think I'm doing this right and I don't have the co-efficient of kinetic friction.
 
Physics news on Phys.org
In the problem it given that a 50 kg mass is moved on a horizontal floor with 250 N. From this data find the coefficient of friction.
 
So, I did:

fk = mu_k*n
250 = mu_k*(50)(9.8)
250/490 = mu_k = .51

So, going back to what I originally did and plugging this back in:

a = [(980)-(50)(9.8)sin15-(.51)(50)(9.8)cos15]/50
a = 12.24 m/s^2

This doesn't seem right, is it?
 
In the calculation, you have the mass as 50 kg. It is not right. It is used to find μ of the surface. For calculation you have to use m and 2m.
Identify the forces acting on m and its acceleration.
Find the acceleration of 2m. And the equate them to find T and a.
 
I'm a bit confused. I thought I had the forces identified in (Fnet)x = n + T - Fgsin\theta - fk = ma, would it be 2ma or?

Sorry, :(.
 
You have to write two equations for two masses.
For m
ma = T -gsinθ -fk. What is the value of fk?
For 2m
2m*a = ...?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Replies
4
Views
2K
Replies
6
Views
9K
Replies
1
Views
2K
Replies
5
Views
2K
Replies
6
Views
4K
Replies
24
Views
9K
Replies
1
Views
5K
Replies
3
Views
2K
Back
Top