A flatbed truck is supported by its four drive wheels, and is moving with an acceleration of 6.3 m/s2. For what value of the coe
1 answer:
Answer:
Coefficient of static friction will be equal to 0.642
Explanation:
We have given acceleration
Acceleration due to gravity
We have to find the coefficient of static friction between truck and a cabinet will
We know that acceleration is equal to , here
is coefficient of static friction and g is acceleration due to gravity
So
So coefficient of static friction will be equal to 0.642
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initially coin is at rest and then it drop for total time t = 1.5 s
so here the speed of the coin at which it will hit the floor is to be find
here we know that
a = 9.8 m/s^2
t = 1.5 s
now from above equation
so it will hit the floor with speed 14.7 m/s
Answer:
Explanation:
mass of pulley, m3 = 1.5 kg
Radius of pulley, R = 0.09 m
mass of monkey, m2 = 4.5 kg
mass of banana bunch, m1 = 3 kg
Let a is teh acceleration ans T1 and T2 be the tension in the rope.
The moment of inertia of the pulley
I = 0.5 x m3 x R² = 0.5 x 1.5 x 0.09 x 0.09 = 0.006075 kgm²
According to Newton's second law
T1 - m1 g = m1 x a .... (1)
m2 g - T2 = m2 x a ..... (2)
(T2 - T1 ) x R = I x α
where, α is the angular acceleration
α = a / R
(T2 - T1)R = 0.5 x m3 x R² x a / R
T2 - T1 = 0.5 x m3 x a ..... (3)
from (1), (2) and (3)
a = 1.78 m/s²
from equation (1)
T1 = m1 ( g + a) = 3 ( 9.8 + 1.78) = 34.77 N
from equation (2)
T2 = m2 (g - a) = 4.5 (9.8 - 1.78) = 36.13 N
