Answer:
176.58 m
Explanation:
t = Time taken = 6 seconds
u = Initial velocity = 0
v = Final velocity
s = Displacement
g = Acceleration due to gravity = 9.81 m/s² = a
Equation of motion
The object travels 176.58 m from the cliff in 6 seconds.
A uniform rod of length 0.8 m and mass 1.4 kg, has two point masses at each end. The point mass on the left end has a mass 1.2 k
1 answer:
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We can solve the problem by requiring the equilibrium of the forces and the equilibrium of torques.
1) Equilibrium of forces:
where
is the weight of the person
is the weight of the scaffold
Re-arranging, we can write the equation as
(1)
2) Equilibrium of torques:
where 3 m and 2 m are the distances of the forces from the center of mass of the scaffold.
Using
and replacing T1 with (1), we find
from which we find
And then, substituting T2 into (1), we find
1) Equilibrium of forces:
where
Re-arranging, we can write the equation as
2) Equilibrium of torques:
where 3 m and 2 m are the distances of the forces from the center of mass of the scaffold.
Using
from which we find
And then, substituting T2 into (1), we find
Answer:
a) T = 2.26 N, b) v = 1.68 m / s
Explanation:
We use Newton's second law
Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string
sin 30 =
cos 30 =
Tₓ = T sin 30
T_y = T cos 30
Y axis
T_y -W = 0
T cos 30 = mg (1)
X axis
Tₓ = m a
they relate it is centripetal
a = v² / r
we substitute
T sin 30 = m (2)
a) we substitute in 1
T =
T =
T = 2.26 N
b) from equation 2
v² =
If we know the length of the string
sin 30 = r / L
r = L sin 30
we substitute
v² =
v² =
For the problem let us take L = 1 m
let's calculate
v =
v = 1.68 m / s