Finding the Critical Angle for Shelf Gliding: A Statics Problem

AI Thread Summary
The discussion revolves around calculating the critical angle at which a bookcase begins to glide rather than tip over when a force is applied. The shelf has a weight of 1500N and a friction force of 600N, with the mass center located at its midpoint. Participants discuss the need to decompose the applied force into horizontal and vertical components to derive a function for the angle. There is confusion regarding the calculations for the forces required to tip and glide the shelf, particularly at different angles. The goal is to establish the minimum angle at which gliding occurs instead of tipping, emphasizing the need for clarity in force analysis.
NathalieSweden
Messages
3
Reaction score
0

Homework Statement



Excuse my english, it is not my first language.
A bookcase has the weight 1500N and is standing at a horizontal floor. The friction force between the floor and the shelf is 600N.
The shelf's height is 1,8mand it is 2m wide. The shelf stands on four feet each 0,01m from the edge. The masscentre is located at the middle of the shelf. A person pulls at the top of one of the shelf's edges. The force creates an angle v with the shelf's side.

I have to form an expression for when the shelf starts gliding, as a function of the angle v.
I will add a file with a pictur that explains the situation.
a2.png

Homework Equations


torque, T=Fxr
Friction, Ff=uN

The Attempt at a Solution


I have calculated the force needed to tip the shelf at v=90degrees: 13500N. Also the force needed to make the shelf start gliding when v=0(paralell)=600N. The force needed to tip the shelf at v=0 is 750N
I have no idea how to form the function.
 
Last edited by a moderator:
Physics news on Phys.org
Oops I posted the image twice...
 
Hello Sweden, welcome to PF :)

My swedish is good enough to see a ##\mu_s=0.4##.
The friction force is ##F_f = \mu_s * F_N##.
##F_N## is the net normal force that compensates mg and the vertical component of the pulling force F
So ##F_f = 600 ## N is only when pulling force F is horizontal (##\theta = {\pi\over 4}## ), which is given in part a).

Were does the 13500 N come from ? Probably part b) 1.9 * 1500 = 0.1 * F
[edit] yes of course, 0.9

ANd you want some function of ##\theta## in part c). That means you have to decompose F in horizontal and vertical components. What kind of function is wanted, my limited swedish doesn't reveal me, so perhaps you want to describe it a bit more extensively ...
 
Last edited:
Yes it comes from b). But it's: 0,1F*0,9*1500. This is correct, the teacher has passed me on a) and b).
Yes that is true. Determine the x and y directed forces for when the shelf glides and tips. Then I am supposed to find the minimum angle where the shelf starts gliding instead of tipping. Do you understand what I mean?
 
"Determine the x and y directed forces for when the shelf glides and tips"

I can't make sense out of that. Can you ? If so, show some steps and indicate where you run into problems...
 
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...
Back
Top