## Homework Statement

A uniform 6.0m long ladder of mass 17.0 kg leans against a smooth wall (so the force F_w exerted by the wall, , is perpendicular to the wall). The ladder makes an angle of 22.0 with the vertical wall, and the ground is rough.Determine the coefficient of static friction at the base of the ladder if the ladder is not to slip when a 75.0kg person stands three-fourths of the way up the ladder.

## Homework Equations

τ=RxF
F=ma
N=normal force from ground

## The Attempt at a Solution

I thought that if I take the sum of torques and the sum of the forces I could determine the static friction but I'm stuck.
Assuming that my axis of rotation is the point where the ladder is leaning against the wall
τ= [1.5 x (75)(9.8)(cos(22))] + [3 x (17)(9.8)(cos(22))] + [Nsin(22)]=0

Y-direction: sum of forces
F=N-(75)(9.8)-(17)(9.8)=0

X-direction: sum of forces
F=µN=0

Hootenanny
Staff Emeritus
Gold Member

## Homework Statement

A uniform 6.0m long ladder of mass 17.0 kg leans against a smooth wall (so the force F_w exerted by the wall, , is perpendicular to the wall). The ladder makes an angle of 22.0 with the vertical wall, and the ground is rough.Determine the coefficient of static friction at the base of the ladder if the ladder is not to slip when a 75.0kg person stands three-fourths of the way up the ladder.

## Homework Equations

τ=RxF
F=ma
N=normal force from ground

## The Attempt at a Solution

I thought that if I take the sum of torques and the sum of the forces I could determine the static friction but I'm stuck.
Assuming that my axis of rotation is the point where the ladder is leaning against the wall
τ= [1.5 x (75)(9.8)(cos(22))] + [3 x (17)(9.8)(cos(22))] + [Nsin(22)]=0

Y-direction: sum of forces
F=N-(75)(9.8)-(17)(9.8)=0

X-direction: sum of forces
F=µN=0

According to your equation, the torque resulting from the frictional force is acting to rotate the ladder in the same direction as the weight of the person and the ladder.

Furthermore, in your equation it appears that the frictional force is acting 1m from the pivot, which is not the case.

According to your equation, the torque resulting from the frictional force is acting to rotate the ladder in the same direction as the weight of the person and the ladder.

Furthermore, in your equation it appears that the frictional force is acting 1m from the pivot, which is not the case.

So, the torque from friction shouldn't be rotating in the same direction as the person?

Which equation do I need to change, the torque?

Hootenanny
Staff Emeritus
Gold Member
So, the torque from friction shouldn't be rotating in the same direction as the person?
In which direction does the frictional force act, towards or away from the wall?

## Homework Statement

A uniform 6.0m long ladder of mass 17.0 kg leans against a smooth wall (so the force F_w exerted by the wall, , is perpendicular to the wall). The ladder makes an angle of 22.0 with the vertical wall, and the ground is rough.Determine the coefficient of static friction at the base of the ladder if the ladder is not to slip when a 75.0kg person stands three-fourths of the way up the ladder.

Actually I am new to this forum and also reading physics after almost 5yrs. So i'm actually trying to solve this problem. So kindly inform me whether my method and answer is correct or not.
I have attached the soln in .bmp format

#### Attachments

40.1 KB · Views: 592
Hootenanny
Staff Emeritus
Gold Member
Actually I am new to this forum and also reading physics after almost 5yrs. So i'm actually trying to solve this problem. So kindly inform me whether my method and answer is correct or not.
I have attached the soln in .bmp format
Welcome to Physics Forums.

Unfortunately, no your solution is not correct. The weight of the person and the ladder does not act through the base of the ladder, it acts through the associated centre of mass. There are only two forces acting at the base of the ladder: the frictional force and the normal reaction force.

Welcome to Physics Forums.

The weight of the person and the ladder does not act through the base of the ladder, it acts through the associated centre of mass. There are only two forces acting at the base of the ladder: the frictional force and the normal reaction force.

So, in that case what force makes the base of the ladder to move away from the wall.
Is the normal force making the ladder to move away from the wall? I think it will be perpendicular to the floor and base of the ladder..... rite??

Hootenanny
Staff Emeritus