What does potential energy depend on?
2 answers:
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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 = (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 =
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
Answer:
The ball would have landed 3.31m farther if the downward angle were 6.0° instead.
Explanation:
In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).
We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.
So, first we need to determine the components of the velocity of the ball, like this:
we pick the positive one, so it takes 0.317s for the ball to hit on point A.
so now we can find the distance from the net to point A with this time. We can find it like this:
Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:
t= -0.9159s or t=0.468s
we pick the positive one, so it takes 0.468s for the ball to hit on point B.
so now we can find the distance from the net to point B with this time. We can find it like this:
So once we got the two distances we can now find the difference between them:
so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.
