How High Can a Person Climb Before a Ladder Tips?

  • Thread starter Thread starter dolpho
  • Start date Start date
  • Tags Tags
    Torque
Click For Summary
SUMMARY

The discussion revolves around a physics problem involving a 5-meter ladder weighing 200 N, positioned against a smooth wall, with a 600 N person climbing it. The key calculations involve determining the forces acting on the ladder, including friction, normal force, and torque. The final solution indicates that the person can climb up to 4.55 meters before the ladder tips, based on a torque balance equation that incorporates the weights of the ladder and the person, as well as the frictional force at the base.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of torque and its calculation
  • Familiarity with frictional forces and coefficients
  • Ability to analyze forces in two dimensions
NEXT STEPS
  • Study the principles of torque and equilibrium in static systems
  • Learn about frictional force calculations in inclined scenarios
  • Explore the application of trigonometric functions in physics problems
  • Practice similar problems involving ladders and tipping points
USEFUL FOR

Students preparing for physics exams, educators teaching mechanics, and anyone interested in understanding static equilibrium and forces in real-world applications.

dolpho
Messages
66
Reaction score
0

Homework Statement



A 5 meter long ladder, weighting 200 N rests against a smooth vertical wall with its base on a horizontal rough floor, a distance of 1.2 meters away from the wall. The center of mass of the ladder is 2.5 m from it's base, and the coefficient of friction is .20. How far up the ladder can a 600 N person climb before the ladder begins to tip.


Homework Equations


Ffriction = uN
F=ma
torque = RF

The Attempt at a Solution



So first I try identifying all of the forces acting on the ladder. In the X direction we have the Frictional force, and the wall acting on the ladder. In the vertical direction we have 200N from the ladder, the 600 N person, and the Normal force acting up on the base of the ladder.

The angle it makes with the wall is 76.1 degrees.

I use Newtons Second law for each of these;

Fy = Ffriction - Fwall = 0
Ffriction = Fwall

Fx = N - 200 - 600 = 0
N = 800 Newtons

Now hopefully I did that part right but when I sum the torques I get confused. So I guess I need to set an axis and then pick which directions are positive or not. I pick the Normal force as my axis. I'll pick counter clockwise as positive and clockwise negative.

Sum of Torques = N(0) - 200 (2.5) - 600d + Fwall(5) = 0

The Fwall is equal to uN = (.2)(800)(cos 76)??

Yea I'm pretty lost here. Any help would be appreciated!
 
Physics news on Phys.org
:cry:
dolpho said:

Homework Statement



A 5 meter long ladder, weighting 200 N rests against a smooth vertical wall with its base on a horizontal rough floor, a distance of 1.2 meters away from the wall. The center of mass of the ladder is 2.5 m from it's base, and the coefficient of friction is .20. How far up the ladder can a 600 N person climb before the ladder begins to tip.


Homework Equations


Ffriction = uN
F=ma
torque = RF

The Attempt at a Solution



So first I try identifying all of the forces acting on the ladder. In the X direction we have the Frictional force, and the wall acting on the ladder.
yes
In the vertical direction we have 200N from the ladder, the 600 N person, and the Normal force acting up on the base of the ladder.
yes
The angle it makes with the wall is 76.1 degrees.
and yet another yes!
I use Newtons Second law for each of these;
well, his first, which is not unlike his 2nd when a = 0.
Fy = Ffriction - Fwall = 0
Ffriction = Fwall

Fx = N - 200 - 600 = 0
N = 800 Newtons

Now hopefully I did that part right
yes, but you have reversed x and y
but when I sum the torques I get confused. So I guess I need to set an axis and then pick which directions are positive or not. I pick the Normal force as my axis. I'll pick counter clockwise as positive and clockwise negative.

Sum of Torques = N(0) - 200 (2.5) - 600d + Fwall(5) = 0
Moment is force times perpendicular distance to the pivot...try that one again.
The Fwall is equal to uN = (.2)(800)(cos 76)??
why no, the friction force is μN, and that is a horizontal force at the base of the ladder, and equal to Fwall...why the cos term?
Yea I'm pretty lost here. Any help would be appreciated!
I think you'll be unlost in a bit, i hope...
 
PhanthomJay said:
:cry: yes yes and yet another yes! well, his first, which is not unlike his 2nd when a = 0. yes, but you have reversed x and y Moment is force times perpendicular distance to the pivot...try that one again. why no, the friction force is μN, and that is a horizontal force at the base of the ladder, and equal to Fwall...why the cos term? I think you'll be unlost in a bit, i hope...

Yea sorry I typed it out wrong. So in the Y direction we have----> N - 200 - 800 = 0
So N=800

in the X we have Ffriction = Fwall

Now for the summing torques. Since the two forces pushing in the clockwise direction, would they have the same angle?

I redid the perpindicular force and hopefully found that its cos(13.9) = 800 / Fperp = 824 n ?

So for the ladder it would be -- > cos(13.9) = 200 / Fperp = 206 n ?

Sum of Torques = -824d - (206)(2.5) + Fwall (5) = 0

The torque of the Fwall would be umm...Fwall / cos13.9 ?

And the Fwall would be equal to the Frictional force since it isn't moving right? So F friction = (.2)(200 Newtons) = 40?

So I guess the total sum of the torques would be...

Sum Torque = -824d - 516 N + 1026N = 0

Dangit I know that's wrong. lol :(

(edit: Do I have all of the correct torques on the equation? So the normal force and Ffrictional force go away since I set is as my axis. So that just leaves the persons torque, the ladders torque and the torque of the wall ?)
 
Last edited:
dolpho said:
Yea sorry I typed it out wrong. So in the Y direction we have----> N - 200 - 800 = 0
So N=800

in the X we have Ffriction = Fwall

Now for the summing torques. Since the two forces pushing in the clockwise direction, would they have the same angle?

I redid the perpindicular force and hopefully found that its cos(13.9) = 800 / Fperp = 824 n ?
At the base of the ladder, you already calculated the vertical reaction to be 800 Newtons. That vertical reaction IS the Normal force at that point. The normal force acts perpendicular to the contact surface (the floor), not along the slope of the ladder.
So for the ladder it would be -- > cos(13.9) = 200 / Fperp = 206 n ?

Sum of Torques = -824d - (206)(2.5) + Fwall (5) = 0
no...
The torque of the Fwall would be umm...Fwall / cos13.9 ?

And the Fwall would be equal to the Frictional force since it isn't moving right? So F friction = (.2)(200 Newtons) = 40?

So I guess the total sum of the torques would be...

Sum Torque = -824d - 516 N + 1026N = 0

Dangit I know that's wrong. lol :(
it is wrong, read on...
(edit: Do I have all of the correct torques on the equation?
no...ooo
So the normal force and Ffrictional force go away since I set is as my axis. So that just leaves the persons torque, the ladders torque and the torque of the wall ?)
Yes this part is more or less correct, when summing torques about the base of the ladder, the vertical normal force and the horizontal friction force of the floor on the ladder have no torque about the base, since the loads pass through that point.

There are 2 issues here: solving for the horizontal and vertical reaction forces of the floor on the ladder at the base of the ladder, and solving for the horizontal reaction force of the wall on the ladder. The vertical normal force is __??___ N and then the limiting friction force is___??___, in which direction? Then once you know the friction force, then the wall force must be equal to it and in which direction?

Then now you should sum torques of the ladder weight, the man's weight, and the wall force, about the ladder base. There are 2 ways to calculate torques. The first is to use the cross product rule T = r X F = rFsintheta, where r is the magnitude of the position vector distance between the point of force application and the point you are summing torques about, and theta is the included angle between the force and position vector. The second way is to use T = force times perpendicular distance from the line of action of the force to the point about which you are summing torques. Let me start you off by calculating the torque about the base of the ladder's weight: Using the first method, T = r X F = rF sintheta = (2.5)(200) sin 13.9. Using the 2nd method, T = F*(perp. distance) = 200(2.5)cos 76.1. Both yield the same result. Watch cw vs. ccw plus or minus signs. It may be a bit easier here to use the first method for the calcs, although the 2nd gives a better understanding of the concept, but with some added geometry/trig. Continue...
 
Ok I think I got this now...

Sum Torques = Fwall(5) - (200N)(2.5) - 600(x) = 0

And then I find the perpendicular forces of the above which is...

Fwall = Sin 76.1 x 800(.2) = 155 N
Ladder force = Sin13.9 x 200(2.5)
Person force = Sin13.9 x 600(d)

Sum Torque = 775 - 120 - 144d = 0
d=4.55 Hopefully that's right but one question I had was when I was trying to find the perpindicular force for the first one, I had a right triangle and I did...

Sin(76.1) = Fperp / 160 = 155

But if I took the cosine wouldn't it be the same answer?

Cos(13.9) = Fwall / 160 = 165?

Sorry if it's a stupid question I always thought that it didn't matter if you took sine or cosine. Thanks for the help!
 
dolpho said:
Ok I think I got this now...

Sum Torques = Fwall(5) - (200N)(2.5) - 600(x) = 0
scratch this equation please it is not correct
And then I find the perpendicular forces of the above which is...

Fwall = Sin 76.1 x 800(.2) = 155 N
Ladder force = Sin13.9 x 200(2.5)
Person force = Sin13.9 x 600(d)

Sum Torque = 775 - 120 - 144d = 0
d=4.55
well I gave you two ways to find torque, and darned you found a third way by finding the perpendicular component of the force to the ladder and using the distance along the ladder, which is ok to do also since the component of the force parallel to the ladder produces no torque.
Hopefully that's right
yes, good work
but one question I had was when I was trying to find the perpindicular force for the first one, I had a right triangle and I did...

Sin(76.1) = Fperp / 160 = 155

But if I took the cosine wouldn't it be the same answer?

Cos(13.9) = Fwall / 160 = 165?
check your calculator again that comes out to 155 same result
Sorry if it's a stupid question I always thought that it didn't matter if you took sine or cosine. Thanks for the help!
no such thing as a stupid question. Sin theta = cos(90 - theta), you just slipped a digit when using your calculator.
 
Oh lol...I had it in radians. WOOPS!

Thank you so much for the help! I appreciate it :) I have my Physics final next week and a test Friday so I might see you again. :D
 

Similar threads

Force of the Ladder on a Wall Torque
  • · Replies 5 ·
Replies
5
Views
3K
How Does Equilibrium Apply in the Classic Ladder Problem?
  • · Replies 3 ·
Replies
3
Views
2K
A ladder against a wall problem
  • · Replies 5 ·
Replies
5
Views
3K
What is the maximum height a man can climb on a ladder before it slips?
  • · Replies 2 ·
Replies
2
Views
3K
Ladder sliding down with no friction
  • · Replies 11 ·
Replies
11
Views
5K
How Do You Calculate Torques for a Ladder in Static Equilibrium?
  • · Replies 2 ·
Replies
2
Views
2K
Static equilibrium -- interpretation of forces
  • · Replies 10 ·
Replies
10
Views
3K
How Much Friction is Needed to Keep the Ladder in Place?
  • · Replies 2 ·
Replies
2
Views
2K
Equilibrium of Forces and Torques on Sawhorses Supporting a Person on a Board
  • · Replies 6 ·
Replies
6
Views
2K
Torque Question: How Far Up a Ladder Can 800N Worker Climb?
  • · Replies 10 ·
Replies
10
Views
2K