• Support PF! Buy your school textbooks, materials and every day products Here!

What is the actual size of the frictional force

  • Thread starter mybrohshi5
  • Start date
  • #1
365
0

Homework Statement



A uniform ladder 5.1 m long rests against a frictionless, vertical wall with its lower end 3.1 m from the wall. The ladder weighs 165 N. The coefficient of static friction between the foot of the ladder and the ground is 0.39 . A painter weighing 739 N climbs slowly up the ladder.

What is the actual size of the frictional force when the painter has climbed 0.9393 m along the ladder?

Homework Equations




The Attempt at a Solution



I am so confused on this type of torque problem. I do not even know where to begin.

this is what i came up with.

I used where the ladder touches the floor as my pivot point.

t = torque

tman + tladder - tfric = 0

165N(0.9393m) + 739N(5.1/2m) - fs(5.1) = 0

i feel like i am leaving some stuff out :(

I am greatly confused and any help would be very appreciated.

Thank you :)
 
Last edited:

Answers and Replies

  • #2
111
0


Treating the foot of the ladder as the pivot point is correct. However, your equation is wrong as it stands. Probably due to a typo, you have the weights of the man and ladder switched around. Also, your final equation will need to make use of the angle, [itex]\theta[/itex], between the ladder and the ground.

As you said, since the ladder experiences no angular acceleration, the net torque acting on it must be zero. You are right in summing the torques due to the weights of the man and ladder, but you need to use [itex]\tau = rF\sin \theta [/itex]. The sum of these two torques is balanced by the single torque acting at the top of the ladder due to the horizontal force of the frictionless wall.

When you start to see (sin/cos) or (cos/sin) terms appearing in your equations then you are probably close to the solution.
 
  • #3
365
0


i am a little confused as to where to use sin and cos. Like sin for which terms and cos for which terms.

so this is what i have right now.

739N(.9393m)(sin(52.57)) + 165N(5.1/2m)(sin(52.57)) - fs(5.1) = 0

fs = 173.6 N

It says my answer is wrong but close. Can you see why?

thanks for the input :)
 
Last edited:
  • #4
365
0


can anyone see why this is wrong? why is it close but not right?

thank you
 
  • #5
365
0


Can anyone please help me with why this is wrong.

My hw is due soon and this is my only problem left :(
 
  • #6
111
0


I always need to sketch out a large diagram for this type of question. Identify similar triangles on the diagram and you will see that your sin(52.57) is not correct, and your fs(5.1) needs to have either a sin(52.57) or a cos(52.57) term in it.

By chance, it seems that the answer to this question is only a few percent lower than your answer. That must be why it is telling you you're close!
 
  • #7
365
0


I ended up figuring out that this was the right equation to use.

[itex]739N(0.9393m)(cos(52.57)) + 165N(\\frac{5.1m}{2})(cos(52.57)) - f_s(5.1m)(cos(37.43)) = 0 [/itex]

[itex]f_s = 167.3 N [/itex]

I am a little confused though when to use sin and cos and what angle to use with them when doing a torque problem like this. Could you explain when i should use sin or cos and when to use what angle.

I am just a little confused about that, but i get the rest of the problem :)

Thank you
 

Related Threads on What is the actual size of the frictional force

  • Last Post
Replies
4
Views
719
Replies
4
Views
2K
  • Last Post
Replies
9
Views
5K
Replies
5
Views
497
Replies
7
Views
4K
Replies
2
Views
1K
Replies
28
Views
257
Replies
4
Views
894
Replies
18
Views
917
Top