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Answer: The center of gravity is 1.1338 m away from the left side of the barbell
Explanation:
Length of the barbell = 1.90 m
The distance center of gravity from left = x
Mass on the left side = 25 kg
The distance center of gravity from right = 1.90 - x
Mass on the right side = 37 kg
At the balance point:
The center of gravity is 1.1338 m away from the left side of the barbell
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]