Statics, ladder leaning against wall finding friction mass etc. really simple q.

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Discussion Overview

The discussion revolves around a physics homework problem involving a uniform ladder leaning against a wall. Participants are exploring the forces acting on the ladder, particularly focusing on the decomposition of the weight into its components and the calculation of the coefficient of friction. The scope includes technical reasoning and mathematical analysis related to statics.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant expresses uncertainty about how to break down the weight of the ladder into its horizontal and vertical components, questioning whether to use sine or cosine for the angle.
  • Another participant suggests that the components of force should be determined based on the direction of the force and the angle involved, indicating that cosine is used for the adjacent side and sine for the opposite side.
  • A different participant advises looking for a point to take moments to derive an equation, noting that many people confuse sine and cosine in such problems and recommending the use of geometrical ratios instead.
  • Another participant clarifies that the force acting against the wall is the normal force, which is perpendicular to the contact point, and provides a breakdown of the weight components as mg sin(60) in the X direction and -mg cos(60) in the Y direction.

Areas of Agreement / Disagreement

Participants express differing views on how to approach the problem, particularly regarding the decomposition of forces and the use of sine and cosine. There is no consensus on the correct method for breaking down the weight components or the best approach to solve the problem.

Contextual Notes

There are unresolved assumptions regarding the definitions of forces and angles, as well as the specific conditions under which the ladder is on the verge of slipping. The discussion does not clarify the mathematical steps necessary to arrive at the coefficient of friction.

Gogarty
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This Is really simple I just can't remember what way Sin and Cosine go

Homework Statement


Here is the problem:
7 (b) A uniform ladder rests on rough
horizontal ground and leans against
a smooth vertical wall.
The length of the ladder is 5 m and
its weight is 80 N.
The angle between the ladder and the
ground is 60
The ladder is on the point of slipping.
(i) Show on a diagram all the forces acting on the ladder.
(ii) Calculate the value of the coefficient of friction


right so I realize the weight acts through the center of gravity in this case half way up. but it also acts towards the wall. is it the sin or cos of the angle that acts down. which goes horizontaly? I know how to do the rest just what way do the components of the weight break up? Is it 80 cos(60 or 80 Sin(60 to find the force acting to the right ?



The Attempt at a Solution

 
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Gogarty said:
This Is really simple I just can't remember what way Sin and Cosine go

Hi Gogarty! Welcome to PF! :smile:

If you're asking about components of force, it's always cos of the angle between the direction of the force and the direction in which you're taking components.

The only time you use sin is when that angle is already called (90º - θ) … so you use cos(90º - θ), which is sinθ. :wink:

If you're asking about moments, use the distance from the point to the line (of force).
right so I realize the weight acts through the center of gravity in this case half way up. but it also acts towards the wall. is it the sin or cos of the angle that acts down. which goes horizontaly? I know how to do the rest just what way do the components of the weight break up? Is it 80 cos(60 or 80 Sin(60 to find the force acting to the right ?

Sorry, I'm not understanding this :redface:

how can weight act towards the wall? …

and why would you want to break the weight into components? :confused:

Just find the normal force, then find the friction force …

what do you get? :smile:
 
Look for a point about which you can take moments to reveal an equation which gives you the answer you want. In my experience, people often get their sins and coses the wrong way round. and I try to avoid them by using geometrical ratios and similar triangles. But try the moment equation first.
 
the force you're talking about acting against the wall is the normal force, and it is perpendicular to the point of contact.

as for the weight, in the X direction it should be mgsin(60) and the Y -mgcos(60)
 

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