Respuesta :
We are given with
l = 8.00 m
θ = 37.0°
m = 470 kg
r = 0.300 m
μ = 0.120
To solve the magnitude of the force required to roll the cylinders up the ramp, we do a force balance
F = Ff
F = μFn
F = mgμ sin θ
substituting the given values
F = 470 (9.81) (0.120) sin 37.0°
F = 333.97 N
l = 8.00 m
θ = 37.0°
m = 470 kg
r = 0.300 m
μ = 0.120
To solve the magnitude of the force required to roll the cylinders up the ramp, we do a force balance
F = Ff
F = μFn
F = mgμ sin θ
substituting the given values
F = 470 (9.81) (0.120) sin 37.0°
F = 333.97 N
Answer:
a) F = 321 N , b) t = 0.87 s
Explanation:
a) For this exercise we will use Newton's second law, in a reference system with an axis parallel to the ramp (x axis) and the other is perpendicular (y axis)
Y Axisy
N - [tex]W_{y}[/tex] = 0
N = [tex]W_{y}[/tex]
X axis
F - fr- Wₓ = 0
F = fr + Wₓ
The expression for the force of friction is
fr = μ N
Let's use trigonometry to find the weight components
sin37 = Wₓ / W
cos37 = [tex]W_{y}[/tex] / W
Wₓ = W sin37
[tex]W_{y}[/tex] = W cos 37
Let's replace
F = μ W cos 37 + W sin37
F = mg (μ cos37 + sin37)
F = 470 9.8 (0.120 cos 37 + sin37)
F = 321 N
b) For this part we must use the relationship between work and energy
W = ΔEm
The work is
W = F L cos 0 = F L
The initial energy at the bottom is
Em₀ = 0
The energy at the top
[tex]Em_{f}[/tex] = K + U = ½ m v² + m g y
Let's look for the height (y) by trigonometry
sin 37 = y / L
y = L sin37
Let's substitute
F L = ½ m v² + m g L sin37 - 0
v² = 2L (F / m - g sin37)
v = √ 2L (F / m - g sin37)
v = √ (2 8 (321/470 - 9.8 sin37) ) = √ ( 16 ( 5.21)
v = 9.13 m / s
As the cylinder rises at a constant speed we can use the ratio
v = d / t
t = d / v
t = L / v
t = 8 / 9.13
t = 0.87 s