Force to raise a cylinder over an edge

• Karol
In summary, the problem involves a cylinder with a mass of 200[kg] and a radius of 0.8[m] attempting to climb a 0.3[m] high edge with a force F exerted at an angle of 30 degrees. The first question asks for the force needed to climb the edge, and the second question asks for the minimum force needed. By making moments around the edge, a true answer of F=1670[N] can be obtained. However, for the second question, the meaning of "minimal force" is unclear and could potentially involve adjusting the angle or point of application of the force. Work-energy theorem is suggested as a possible solution, but it leads to an incorrect answer.
Karol

Homework Statement

The cylinder with mass 200[kg] and radius 0.8[m] tries to climb a 0.3[m] high edge with a force F exerted at an angle of 300. what's the force and what's the minimum force for the cylinder to climb

Homework Equations

Moments: F1L1=F2L2

The Attempt at a Solution

The weight balances the the force, i made moments around the edge and got a true answer F=1670[N], but i don't understand the second question, about the minimum force, and it's about 766[N]

Attachments

• Snap1.jpg
10 KB · Views: 426
"Minimal force" might mean the force needed if the angle can be adjusted. But then the two values should not be so different. Maybe you can adjust the point where it acts on as well. If that is the full problem statement and 766 N is the given answer, it is poorly written.

mfb said:
"Minimal force" might mean the force needed if the angle can be adjusted. But then the two values should not be so different. Maybe you can adjust the point where it acts on as well. If that is the full problem statement and 766 N is the given answer, it is poorly written.
To get an answer so much lower, it must be allowed to apply the force further from the step.

mfb said:
Maybe you can adjust the point where it acts on as well.

haruspex said:
To get an answer so much lower, it must be allowed to apply the force further from the step.
Ah! Yes. That's how you can get about 766 N.

Last edited:
Hi Karol.

I consider that the ball pivots about point of contact (O). So the CoM of the ball will perform circular motion about point O.

Now - work energy theorem states that 'Work done by all the forces on the system is equal to the change in the kinetic energy'.
As the required force is minimum so the ball the velocity of the CoM of the ball at the final position (i.e when the CoM is just above the point O) should be zero.

Now there are three force acting on the ball -- gravity, F and Normal force (N). Work done by N is zero.

Work done by F=work done by mg...(1)

Work done by mg can be find easily.

The black curve shows the trajectory of the CoM. Angle ##\phi## can be found easily. Let the Com be at any arbitrary position such that it has rotated by angle ##\theta## about O.

Work done by F is ##\int _{ 0 }^{ (90-\phi ) }{ FRcos(30-(\theta +\phi ))d\theta } ##.

Just put the value in eq. (1). I think you will get the answers.

Last edited:
Work is either torque times angular displacement of force times displacement and in both ways i get an other equation than: ##\int _{ 0 }^{ (90-\phi ) }{ FRcos(\theta +\phi )d\theta }##
According to the new drawing, for the initial position: α=90-(30+Φ) so W=FR cos[90-(30+Φ)]
Of course we add θ for an arbitrary position and take an integral at the same boundaries

Attachments

• Snap1.jpg
9.6 KB · Views: 436
Last edited:
Karol said:
Work is either torque times angular displacement of force times displacement and in both ways i get an other equation than: ##\int _{ 0 }^{ (90-\phi ) }{ FRcos(\theta +\phi )d\theta }##
According to the new drawing, for the initial position: α=90-(30+Φ) so W=FR cos[90-(30+Φ)]
Of course we add θ for an arbitrary position and take an integral at the same boundaries

I made a mistake in #post 5.
The work done by F is ##\int _{ 0 }^{ (90-\phi ) }{ FRcos((\theta +\phi)-60 )d\theta }##

If you consider the AO as the initial condition and BO as the final condition you will find that initially ##\theta=0## and final ##\theta=90-\phi##.

What would you get after evaluating the integral?

@Satvik Pandey: That does not work. What you calculate would be the average force needed, but the force depends on the angle, so you underestimate the force needed to start the lifting process.

Satvik Pandey
The geometry i have drawn is to scale, the force is above the tangent, at the initial position A and further, not like you drew, so the integral is:
$$W=\int_0^{51.32} F\cdot 0.8 \cdot \cos (60-\phi)d\phi=-0.8F\cdot\sin(60-\phi)\vert^{51.32}_0=-0.8F(\sin 8.68-\sin 60)=0.57F$$
The gain in potential enrgy equals W: ##0.57F=2000[N]\cdot 0.3[m]\rightarrow F=1053##
But that is only theoretical since it's smaller than the force needed to lift it at the beginning

Karol said:
The geometry i have drawn is to scale, the force is above the tangent, at the initial position A and further, not like you drew, so the integral is:
$$W=\int_0^{51.32} F\cdot 0.8 \cdot \cos (60-\phi)d\phi=-0.8F\cdot\sin(60-\phi)\vert^{51.32}_0=-0.8F(\sin 8.68-\sin 60)=0.57F$$
The gain in potential enrgy equals W: ##0.57F=2000[N]\cdot 0.3[m]\rightarrow F=1053##
But that is only theoretical since it's smaller than the force needed to lift it at the beginning
As mfb has pointed out, it is not only unnecessary to consider energy or perform integrals to answer this question, it will give the wrong answer.
All you have to do is figure out where and at what angle to apply the force to get the cylinder moving. Beyond that, you could consider whether a larger force may be required at some later point in lifting it over the step (if the direction and point of application are not allowed to change?).

mfb said:
@Satvik Pandey: That does not work. What you calculate would be the average force needed, but the force depends on the angle, so you underestimate the force needed to start the lifting process.

Thank you mfb for pointing it. Sorry Karol, I lead you on the wrong path.:s

What is the definition of "force to raise a cylinder over an edge"?

The force to raise a cylinder over an edge refers to the amount of energy or effort required to lift a cylinder up and over a surface or barrier, such as an edge or lip.

What factors affect the force to raise a cylinder over an edge?

The force to raise a cylinder over an edge is affected by several factors, including the weight of the cylinder, the height and angle of the edge, and the friction between the cylinder and the surface it is being raised over.

How can the force to raise a cylinder over an edge be calculated?

The force to raise a cylinder over an edge can be calculated using the formula F = mg + μmg, where F is the force, m is the mass of the cylinder, g is the acceleration due to gravity, and μ is the coefficient of friction between the cylinder and the surface.

Can the force to raise a cylinder over an edge be reduced?

Yes, the force to raise a cylinder over an edge can be reduced by decreasing the weight of the cylinder, decreasing the height or angle of the edge, or by using a lubricant to reduce friction between the cylinder and the surface.

How does the force to raise a cylinder over an edge relate to real-world applications?

The concept of force to raise a cylinder over an edge is important in various real-world applications, such as moving heavy objects over a curb or step, lifting objects onto a higher surface, and designing machinery that requires lifting or moving objects over edges or obstacles.

• Introductory Physics Homework Help
Replies
1
Views
730
• Introductory Physics Homework Help
Replies
11
Views
1K
• Introductory Physics Homework Help
Replies
16
Views
2K
• Introductory Physics Homework Help
Replies
22
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
3K
• Introductory Physics Homework Help
Replies
1
Views
843
• Introductory Physics Homework Help
Replies
5
Views
6K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
4K
• Introductory Physics Homework Help
Replies
47
Views
7K