Answer:
μ = 0.37
Explanation:
For this exercise we must use the translational and rotational equilibrium equations.
We set our reference system at the highest point of the ladder where it touches the vertical wall. We assume that counterclockwise rotation is positive
let's write the rotational equilibrium
W₁ x/2 + W₂ x₂ - fr y = 0
where W₁ is the weight of the mass ladder m₁ = 30kg, W₂ is the weight of the man 700 N, let's use trigonometry to find the distances
cos 60 = x / L
where L is the length of the ladder
x = L cos 60
sin 60 = y / L
y = L sin60
the horizontal distance of man is
cos 60 = x2 / 7.0
x2 = 7 cos 60
we substitute
m₁ g L cos 60/2 + W₂ 7 cos 60 - fr L sin60 = 0
fr = (m1 g L cos 60/2 + W2 7 cos 60) / L sin 60
let's calculate
fr = (30 9.8 10 cos 60 2 + 700 7 cos 60) / (10 sin 60)
fr = (735 + 2450) / 8.66
fr = 367.78 N
the friction force has the expression
fr = μ N
write the translational equilibrium equation
N - W₁ -W₂ = 0
N = m₁ g + W₂
N = 30 9.8 + 700
N = 994 N
we clear the friction force from the eucacion
μ = fr / N
μ = 367.78 / 994
μ = 0.37
Answer:
The object will not move.
Explanation:
If nothing pushes against it it will not move. If its not on a slant it will not move.
Transverse, I think. I may be wrong.
While riding in a hot air balloon,
which is steadily at a speed of 1.01 m/s, and your phone accidentally falls.
<span>(a)
</span>The
speed of your phone after 4 s is:
V= u +
at
V= 1.01
+ (9.8)(4)
V=
40.21 m/s
<span>(b)
</span>The balloon
is ____ far:
V = u +
at
V= 1.01
+ (9.8)(1)
V=10.81
–distance at 1 one second
V= u +
at
V= 1.01
+ (9.8)(2)
V= 20.61-distance
at 2 seconds
V= u+ at
V=
30.41- distance at 3 seconds
V=
40.21- distance at 4 seconds
D=
102.04 m
<span>(c)
</span>If the
balloon is rising steadily at 1.01 m/s:
V= -1.1
m/s
<span> </span>
