For this exercise we use the rotational equilibrium condition, we have a reference system is the floor and the anticlockwise rotations as positive, in the adjoint we can see a diagram of the forces

let's use subscript 1 for the ladder and 2 for the firefighter

∑ τ = 0

-W₁ x₁ - W₂ x₂ + N₁ y = 0

N₁ = $\frac{W_1 x_1 + W_2 x_2}{y}$ (1)

the center of mass of the ladder is at its geometric center,

d = L / 2 = 15/2 = 7.5 m

cos 60 = x₁ / d₁

x₁ = d₁ cos 60

x₁ = 7.5 cos 60

x₁ = 3.75 m

for the firefighter d₂ = 4 m

cos 60 = x₂ / d₂

x₂ = d₂ cos 60

x₂ = 4 cos 60 = 2 m

for the fulcrum d₃ = 15 m

sin 60 = y / d₃

y = d₃ sin 60

y = 15 sin 60

y = 13 m

we look for the Normal by substituting in equation 1

N₂ = $\frac{500 \ 3.75 \ + 800 \ 2}{13}$

N₂ = 267.3 N

now let's use the translational equilibrium relations

X axis

F₁ - N₂ = 0

F₁ = N₂

F₁ = 267.3 N

Axis y

N₁ - W₁ -W₂ = 0

N₁ = W₁ + W₂

N₁ = 500 + 800

N₁ = 1300 N

b) for this case change the firefighter's distance d₂ = 9 m

x₂ = 9 cos 60

x₂ = 4.5 m

we substitute in 1

N₂ = \frac{500 \ 3.75 \ + 800 \ 4.5}{13}

N₂ = 421.15 N

of the translational equilibrium equation on the x-axis

fr = F₁ = N₂

fr = 421.15 N

friction force has the expression

fr = μ N

in this case the reaction of the Earth to the support of the ladder is N1 = 1300N

μ = fr / N₁

μ = 421.15 / 1300

μ = 0.324

8 0

2 years ago

A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-

alexdok [17]

Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, $q_1=-3\ nC$

It is placed at a distance of 9 cm at x axis

Charge, $q_2=+4\ nC$

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

$\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0$

Here,

$r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}$

So,

$\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}$

Squaring both sides,

$3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m$

So, at a distance of 10.2 m on the y axis the electric potential equals 0.

8 0

3 years ago