Question 9
1 answer:
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Answer:
x = 25 / μ [ ft]
Explanation:
To solve this exercise we can use Newton's second law.
Let's set a reference system where the x axis is parallel to the road
Y axis
N_B + N_A - W_van - W_load = 0
N_B + N_A = W_van + W_load
X axis
fr = ma
a = fr / m
the total mass is
m = (W_van + W_load) / g
the friction force has the expression
fr = μ N_{total}
fr = μy (W_van + W_load)
we substitute
a = μ (W_van + W_load)
a = μ g
taking the acceleration let's use the kinematic relations where the final velocity is zero
v² = v₀² - 2 a x
0 = v₀² -2a x
x =
x =
x =
x = 25 / μ [ ft]
Answer:
F = 84.61 N
Explanation:
As in the figure, since there is no friction so if component of Force applied along the incline is greater than the component of weight along the incline, then the object will move up the incline.
component of Force along the incline = F cos(23° - 15°) = F cos(8°)
component of weight along the incline = 33*g*sin(15°) = 33*9.81*sin(15°)
Equating the above two components of forces will give the minimum Force required.
F cos(8°) = 33*9.81*sin(15°)
F = 33*9.81*sin(15°) / cos(8°) (calculate the value using a scientific calculator)
<u>F = 84.61 N</u>
