Statics (frictional force) problem

  • Thread starter jakeward46
  • Start date
  • #1
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Homework Statement



The question is:

A uniform ladder of mass of 50kg is palced with its foot on a horizontal floor and its top against a vertical wall. Determine the minimum value of themass which must be placed against the foot of the ladder to retain it at an angle of 45degrees to the floor if the coefficient of friction between all relevant surfaces is 0.2.

Homework Equations



Fs=μN
g=9.81
ladder length = L, and eighter side of mass L/2( in equations)
Joint A=ladder contact with wall
Joint B=ladder contact with floor(supported by mass)

The Attempt at a Solution



I thought the best way was to split the ladder from the mass keeping it upright when summing the forces...

Joint A
∑fy=0=P+Fs-50.g.(L/2)cos45

∑fx=0=N-50.g(sin45)=o......N= 347N
Fs=μN=0.2x347=69.4

then i decided to take moments

M.A=0=50.g.L/2.cos45

Joint B
∑fy=o=N-50.g.sin45=o
N=347N

∑fx=0= -Fs-P-50.g.cos45=0
P= -Fs-347=-416.4N

I dont know where to go with it next as i cnnot get the answer that was provided of m=58.4kg

Any help would be much appriciated. Thanks.
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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251
welcome to pf!

hi jakeward46! welcome to pf! :wink:
Joint A …
Joint B …

i don't follow what you're doing :redface:

the ladder is rigid, you can't split it into joints …

do x components for all the forces on the ladder,

then y components for all the forces on the ladder,

then moments about one end of the ladder (it doesn't matter which) …

what do you get? :smile:
 
  • #3
15
0
Hi tiny-tim, thanks for the reply.

i didnt mean split the ladder, having the imaginary mass at the bottom ladder supporting it, the mass being the figure i need to find.

I was splitting the mass from the ladder but at first i tried just summing the forces on the ladder as you advised but the reactions to the wall and the floor i seem to get too many unknowns?? I will try again :)
 
  • #4
15
0
can anyone help me with this, cannot find anything to help me out with the problem
 
  • #5
741
27

Joint A
∑fy=0=P+Fs-50.g.(L/2)cos45



Let's start with your first equation:
It begins with a P and a Fs as forces, but what does the 50.g.(L/2)cos 45 represent, and what would its units be?

Take the suggestion already given you by tiny tim and draw a picture of the ladder alone. Now put on it all the forces acting on it. Now make equilibrium statements about what you have in front of you.
 

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