Auniform ladder of length l and mass m leans against a frictionless vertical wall, making an angle of 54° with the horizontal. the coefficient of static friction between the ladder and the ground is 0.32. if your mass is four times that of the ladder, what percentage of the way up the ladder can you climb before the ladder begins to slip?

h=0.425 L

Explanation:

Given that

θ =  54°

Coefficient of friction μ = 0.32

Mass of rod = m

Lets take mass of man = M  = 4 m

C is the center of mass of the rod.

By balancing force in y and x direction

R= Fr

R = Fr=  μ N

N = mg + Mg = mg + 4 m g                     ( M =4m)

N = 5 m g

Lets take distance cover by man is h along rod before sliding

Now taking moment about the lower end

M g  h cosθ + m g cosθ L/2  =  R L sinθ

2 M g  h cosθ + m g cosθ L  = 2 R L sinθ

Now by putting the value of R  and M

8 m g  h cosθ + m g cosθ L  = 2  μ N L sinθ

8 m g  h cosθ + m g cosθ L  = 10 m g μ  L sinθ

8   h cosθ +  cosθ L  = 10  μ  L sinθ

8 h + L = 10 μ L tanθ

Now putting the value of θ  and μ

8 h + L = 10 x 0.32 x tan54° x L

8 h + L = 4.4 L

8 h = 3.4 L

h=0.425 L

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.

do you have a formula possibly