Help with force and friction on a flat surface

AI Thread Summary
To determine the minimum force needed to prevent a 30kg crate from moving under a 125N pull, the relationship between static friction and normal force must be understood. The static friction force can be calculated using the coefficient of static friction (0.34) multiplied by the normal force. The normal force is equal to the weight of the crate (294.3N) plus any additional force applied downward. To stop the crate from moving, the total static friction must equal the pulling force, which means the normal force must be adjusted accordingly. Understanding this relationship is crucial for solving the problem effectively.
Mattches
Messages
2
Reaction score
0

Homework Statement


I have a 30kg crate on a flat floor. There is a rope pulling the crate to the right with a force of 125N. The coefficients of static and kinetic friction are .34 and .24 respectively. I need to know what the minimum force I could put on top of the crate to stop it from moving.
Fg = 9.81 m/s^2
uS = .34
uK = .24
m = 30kg


Homework Equations


I know that I need to use Newtons second law to determine the sum of all forces in each direction, and see how much needs to be added to overcome the 125N pull. But honestly I have no idea where to start!


The Attempt at a Solution


I know that Fg on the crate is 294.3N. And I know that friction opposes the way the crate is being pulled. But I really don't know how to piece this problem together.
Can anyone help?
 
Physics news on Phys.org
What must the total static friction be to keep the crate from moving? (Assuming it is initially at rest.)

How is max static friction related to the normal force between crate and floor?
 
The total static friction has to be equal to the force pulling the crate. The normal force is pushing the crate up at the same rate that gravity is pulling down but...I'm still not exactly sure where to go from there. Sorry but I'm very new at this :[
 
Mattches said:
The total static friction has to be equal to the force pulling the crate.
Good.
The normal force is pushing the crate up at the same rate that gravity is pulling down but...
That would be true if you didn't apply any additional force on the crate. But whatever force you push down on the crate with will add to the normal force.

How does static friction relate to the normal force? Use that fact (and your first statement in this post) to figure out what the total normal force needs to be for there to be enough friction.
 
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