Answer:
Explanation:
Assuming x direction is horizontal any y direction is vertical
As the moment of inertia of the pulley is I = (½)mR²
The pulley mass will appear as ½ its actual mass in the system.
F = ma
5.00(9.81) = (5.00 + 2.60/2 + 12.0)a
a = 49.05 / 18.3
a = 2.6803278...
a = 2.68 m/s²
A) T = 12.0(2.68) = 32.16... = 32.2 N
B) T = 5.00(9.81 - 2.68) = 35.65 = 35.7 N
C) aₓ = 2.68 m/s²
D) Fx = 32.2 N
Fy = 35.7 + 2.60(9.81) = 61.2 N
A)The tension in the horizontal segment of the wire will be 32.2 N
B)The tension in the vertical segment of the wire will be 35.7 N
Acceleration is defined as the rate of change of the velocity of the body. Its unit is m/sec².It is a vector quantity. It requires both magnitudes as well as direction to define.
The given data in the problem is;
m is the mass box resting on a horizontal =12.0 kg
the uniform solid disk of mass of 2.60 kg
d is the diameter = 0.500 m.
A)The tension in the horizontal segment of the wire will be 32.2 N
The acceleration is found as;
[tex]\rm F = ma \\\\ a= \frac{49.05}{18.3} \\\\ \rm a= 2.68 \ m/sec^2[/tex]
The tension in the horizontal segment of the wire is found as;
T=ma
T= 12.0 ×2.68
T=32.2 N
Hence the tension in the horizontal segment of the wire will be 32.2 N
B)The tension in the vertical segment of the wire will be 35.7 N
T=m(g-a )
T= 5.00 × (9.8-2.68)
T=35.7 N
Hence the tension in the vertical segment of the wire will be 35.7 N
C) the x-component of the acceleration of the box. will be 2.68 m/s².
aₓ = 2.68 m/s²
D)The x- and y-components of the force that the axle exerts on the pulley will be 32.2 and 61.2 N respectively.
Fx = 32.2 N
Fy = 35.7 + 2.60(9.81) = 61.2 N
Hence the x- and y-components of the force that the axle exerts on the pulley will be 32.2 and 61.2 N respectively.
To learn more about acceleration refer to the link;
https://brainly.com/question/969842
