Question

A ladder of length L is placed against a smooth wall such that it forms a angle θ with the wall and the friction force between the ladder and the floor is fs .

a. On your paper, derive an expression for the ladder's mass, m, in terms of θ, g, fs, and/or L.

b. Try out your derivation. If θ=32° and the friction force on the floor is 47 N, what is the mass of the ladder?

Answer 1

a) m = 2fs(tanθ)/g

b) m = 5.99 kg

Step-by-step explanation:

a) Expression for the ladder's mass, m, in terms of θ, g, fs, and/or L.

Data

m₁ : mass of the lader

g: acceleration due to gravity

L : ladder length

θ : angle that makes the ladder with the floor

µ = 0 : coefficient of friction between the ladder and the wall

fs : friction force between the ladder and the floor

Forces acting on the ladder

W =m*g : Weight of the ladder (vertical downward) , m: mass of the lader

FN :Normal force that the floor exerts on the ladder (vertical upward) (point A)

fs : friction force that the floor exerts on the ladder (horizontal to the left) (point A)

N : Forces that the wall exerts on the ladder (horizontal to the right)

Equilibrium of the forces in X

∑Fx=0

N -fs = 0

N = fs

The equilibrium equation of the moments at the point contact point of the ladder with the floor:

∑MA = 0

MA = F*d

Where:

∑MA : Algebraic sum of moments in the the point (A) (contact point of the ladder with the wall)

MA : moment in the point A ( N*m)

F : Force ( N)

d :Perpendicular distance of the force to the point A ( m )

Calculation of the distances of the forces at the point A

d₁ = (L/2)*cosθ : Distance from W to the point A

d₂ = L*sinθ : Distance from N to the point A

Equilibrium of the moments at the point A

∑MA = 0

N(d₂)-W(d₁) = 0

W( (L/2)*cosθ)= N(L*sinθ )

mg( (L/2)*cosθ)= fs(L*sinθ )

We divided by L both sides of the equation

mg (cosθ/2) =fs(sinθ)

m=2fs(sinθ)/ g( cosθ)

m = 2fs(tanθ)/ g

b) If θ=32° and the friction force on the floor is 47 N, what is the mass of the ladder?

m = 2fs(tanθ)/ g

m = 2(47)(tan32°)/(9,8)

m = 5.99 kg

Answer 2

a.
`m=(2f_(s)tan\theta)/(g)`

b. m=5.99kg

Step-by-step explanation:

a.

In order to solve this problem, we can start by drawing a diagram of the situation. Drawing a diagram is really important since it will help use understand the problem better and analyze it as well. (See attached picture).

In the diagram we can see the forces that are acting on the ladder. We will assume the ladder is static (this is it doesn't have any movement) and analyze the respective forces in the x-direction and the forces in the y-direction, as well as the moments about point B.

So we start with the sum of forces about y, so we get:

`\sum F_(y)=0`

N-W=0

N=W

N=mg

Next we can do the sum of forces about x, so we get:

`\sum F_(x)=0`

which yields:

`f_(s)-F_(W)=0`

so:

`f_(s)=F_(W)`

Next the torque about point B, so we get:

`\sum M_(B)=0`

so:

`f_(s) L sin\theta - NLcos\theta + (WL)/(2)cos\theta = 0`

From the sum of forces in the y-direction we know that N=mg (this is because the wall makes no friction over the ladder) so we can directly substitute that into our equation, so we get:

`f_(s) L sin\theta - WLcos\theta + (WL)/(2)cos\theta = 0`

We can now combine like terms, so we get:

`f_(s) L sin\theta -(WL)/(2)cos\theta = 0`

we know that W=mg, so we can substitute that into our equation, so we get:

`f_(s) L sin\theta -(mgL)/(2)cos\theta = 0`

which can now be solved for the mass m:

`(mgL)/(2)cos\theta = f_(s) L sin\theta`

If we divided both sides of the equation into L, we can see that the L's get cancelled, so our equation simplifies to:

`(mg)/(2)cos\theta = f_(s)sin\theta`

we can now divide both sides of the equation into g so we get:

`(m)/(2)cos\theta = (f_(s)sin\theta)/(g)`

next we can divide both sides of the equation into cos θ so we get:

`(m)/(2)= (f_(s)sin\theta)/(g cos\theta)`

and finally we can multiply both sides of the equation by 2 so we get:

`m=(2f_(s)sin\theta)/(g cos\theta)`

we know that:

`tan \theta=(sin \theta)/(cos \theta)`

so we can simplify the equation a little more, so we get:

`m=(2f_(s)tan \theta)/(g)`

b.

So now we can directly use the equation to find the mass of the ladder with the data indicated by the problem:

θ=32° and
`f_(s)=47N`

we also know that
`g=9.8m/s^(2)`

so we can use our equation now:

`m=(2f_(s)tan \theta)/(g)`

so we get:

`m=(2(47N)tan (32^(o)))/(9.8m/s^(2))`

which yields:

m=5.99kg