Force required to move a object on slope with rolling tyres

AI Thread Summary
To calculate the force required to move a 5-kilogram object on a 30-degree slope with rolling tyres, the formula F push = mgsin(theta) + coefficient of friction * mgcos(theta) was applied, yielding a result of 25.542 Newtons. The coefficient of friction used was 0.001, which is typical for rubber on steel, but there is uncertainty regarding its accuracy due to variations in tyre design and inflation pressure. The discussion highlights that rolling resistance is influenced by multiple factors, including mechanical friction in the wheel/axle system, which complicates the determination of a precise coefficient. While the method of calculation is sound, the lack of a specific rolling resistance value in the problem statement raises questions about the accuracy of the results. Overall, the approach is valid, but further clarification on the coefficient of rolling resistance is necessary for precise calculations.
adssd
Messages
3
Reaction score
0
Consider a object of 5 kilos which is fixed with 4 rolling tyres (like in a trolly) which is in a slanting position of 30 degree angle.

My question:How to find the force required to move the object upwards the slope along with rolling resistance considered?

I tried the force using the F push formula for slope & i tried the coeffiecient for friction of tyres as 0.001

F push= mgsin(theta) + coeff of friction*mgsin(theta)
F push=5 * 9.8 *sin(30) + 0.001 * 5*9.8 * cos(30)
F push=25.542 Newton

Is this correct?

I researched this problem in google, but since the above problem deals with Rolling friction as its attached with tyres, what is the correct method to find the exact force! I s it the correct method to find the object moving up the slope with tyres ?
 
Physics news on Phys.org
You equation is correct.
 
JBA said:
You equation is correct.

is it the correct formulae to use it to my problem?
 
If the above is a full statement of the problem as presented to you then yes it is; however, I am surprised that you were not given a specific coefficient of rolling resistance as a part of the problem statement.
 
JBA said:
If the above is a full statement of the problem as presented to you then yes it is; however, I am surprised that you were not given a specific coefficient of rolling friction as a part of the problem statement.
as i am using steel surface and rubber tyres, i thought its coefficient is 0.001, but i seriously have a doubt over that coeffienct part..as the slope is steel and the tyre is rubber! is the co efficient i used here is right?
 
That is hard to say because the actual rolling resistance of a rubber tyre is a function of the tyre's individual design and even more importantly, its inflation pressure; as a result, to the best of my knowledge there is no one "accepted" value of rolling resistance for inflated rubber tyres.

As for steel wheels or hard rubber covered steel wheels, I have no background related to those; but, in all of these cases the ultimate rolling resistance is a factor that actually must include the mechanical wheel/axle bearing friction, etc. That is the reason, I am surprised you have not been given a specific value in your problem statement.

Ultimately, your equation provides for all possibilities, including a value of "0" so for that reason I would find it hard for anyone to question your method of calculation with no factor value given.
 

Thread '1:1 versus 2:1 Mechanical Advantage debate'

The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
View full post »

Thread 'Average velocity as a weighted mean'

Hello everyone, Consider the problem in which a car is told to travel at 30 km/h for L kilometers and then at 60 km/h for another L kilometers. Next, you are asked to determine the average speed. My question is: although we know that the average speed in this case is the harmonic mean of the two speeds, is it also possible to state that the average speed over this 2L-kilometer stretch can be obtained as a weighted average of the two speeds? Best regards, DaTario
View full post »

Thread 'Newton's first law?'

Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
View full post »

Similar threads

Force required to start rolling a wheel ?
Replies
14
Views
22K
B Questions about momentum and force as well as normal force/friction
Replies
35
Views
3K
Rolling Pendulum: Solving the Dynamics Equations
Replies
2
Views
3K
Direction of friction acting on a rolling object
Replies
12
Views
10K
Rolling down a slope vs to slide down a slope
Replies
36
Views
22K
How can we model a rolling truck on an incline using rotational motion analysis?
Replies
24
Views
2K
Rolling without slipping problem
Replies
42
Views
3K
What is the relationship between torque, linear force, and rolling resistance?
Replies
13
Views
2K
Rolling down a sphere: slipping vs. separating
Replies
3
Views
2K
Wooden sphere rolling on a double metal track
2
Replies
97
Views
5K

Hot Threads

Recent Insights

Back
Top