Which vector should be negative?

A 15 m uniform ladder weighing 500 N rests against a frictionless wall. The ladder makes a 60° angle with horizontal. (a) Find t

scoray [572]

Answer:

a) F₁ = 267.3 N, N₁ = 1300 N, b) μ = 0.324

Explanation:

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 baseball pitcher throws a fastball at 42 meters per second. if the batter is 18 meters from the pitcher, approximately how muc

lapo4ka [179]

T=D/v = 18/42 = 43 seconds

6 0

3 years ago

Read 2 more answers

The particle, initially at rest, is acted upon only by the electric force and moves from point a to point b along the x axis, in

bezimeni [28]

1) Potential difference: 1 V

2) $V_b-V_a = -1 V$

Explanation:

1)

When a charge moves in an electric field, its electric potential energy is entirely converted into kinetic energy; this change in electric potential energy is given by

$\Delta U=q\Delta V$

where

q is the charge's magnitude

$\Delta V$ is the potential difference between the initial and final position

In this problem, we have:

$q=4.80\cdot 10^{-19}C$ is the magnitude of the charge

$\Delta U = 4.80\cdot 10^{-19}J$ is the change in kinetic energy of the particle

Therefore, the potential difference (in magnitude) is

$\Delta V=\frac{\Delta U}{q}=\frac{4.80\cdot 10^{-19}}{4.80\cdot 10^{-19}}=1 V$

2)

Here we have to evaluate the direction of motion of the particle.

We have the following informations:

- The electric potential increases in the +x direction

- The particle is positively charged and moves from point a to b

Since the particle is positively charged, it means that it is moving from higher potential to lower potential (because a positive charge follows the direction of the electric field, so it moves away from the source of the field)

This means that the final position b of the charge is at lower potential than the initial position a; therefore, the potential difference must be negative:

$V_b-V_a = - 1V$