Friction coefficient multiple choice and ladder problem

[cracktheegg]

Homework Statement

Question 1


Which statement is true of the coefficient of static friction?
Answer

It is the force resisting motion.

It is proportional to the static friction force.

It is a dimensionless quantity expressing the ratio of limiting friction to normal reaction.

It is the ratio of friction to normal reaction.

Question 2


The ratio of limiting friction to the normal reaction is known as
Answer

angle of friction.

angle of repose.

coefficient of static friction.

coefficient of kinetic friction.


Question 3


Which of the following statements is NOT true about friction?
Answer

Kinetic friction is larger than static friction.

Limiting static friction is dependent on normal reaction acting on the body.

Coefficient of friction denotes surface roughness of the bodies in contact.

Frictional force acts tangential to the surfaces in contact.

Problem sum

https://www.icloud.com/photostream/#AE5oqs3qHTNWQ;6FA15CFC-95B7-4C76-83AF-405A75919F01

Homework Equations

The Attempt at a Solution

1. 3rd option
2. 3rd option
3. 1st option.

Problem sums:

Where W is the skier's weight:
Force pressing skier agaist slope = W×Cosθ
Friction force in motion = 0.07×W×Cosθ ... (1)
Skier's weight component acting down the slope = W.Sinθ ... (2)
(1) and (2) are equal giving:
0.07×W×Cosθ = W.Sinθ from which:
Tanθ = 0.07 so that θ = arc-tan(0.07) = 4.0˚


Weight of ladder = 18×9.8 = 176.4 N
Weight of man = 90×9.8 = 882.0 N
Put the man at the top of the ladder
Vertical component of force between ladder and ground (normal force) =Fn= 174.6 + 882 = 1058.4 N
To find reaction of wall against the ladder take moments of all applied forces about bottom end of ladder:
(176.4×2 + 882×4).sin(30˚) = Rw×4.Cos(30˚) (Rw = reaction of wall)
Rw = (176.4×2 + 882×4).sin(30˚) / [4×Cos(30˚)]
(176.4×2 + 882×4).½ /(2.√3) = 560.2 N
The horizontal component of the reaction force between ladder and ground = Fh = Rw since all horizontal forces on the ladder must balance and there are only these two!
Necessary friction coefficient = Fh/Fn = 560.2/1058.4 = 0.529

 

[nschaefe]

It all looks right to me.

You should note that in the second problem, you can more easily (in my opinion) retrieve the coefficient of static friction by taking moments about the top of the ladder instead of the bottom.

$\sum M_{t} = 0 = W_{L}*(L/2)*sin(30) - F_{y}*L*sin(30) + F_{x}*L*cos(30)$

$\sum F_{y} = 0 = F_{y} - W_{L} - W_{M}$

$F_{x} = μF_{y}$

where $F_{y}$ and $F_{x}$ act on the bottom of the ladder

Solving for $F_{y}$ and $F_{x}$ in the force and friction equations, them plugging into the moment equation, you can solve for $μ$ more easily

 

[cracktheegg]

It seem that the ski question answer may be wrong according to the result i get, any clue why?

 

[PhanthomJay]

It seem that the ski question answer may be wrong according to the result i get, any clue why?

Looks ok to me...