Respuesta :
Answer:
Explanation:
a )
Since pulley is friction-less , tension on both the sides of pulley will be same.
Let it be T . The pulley will not rotate because it is friction-less.
For motion of 5 kg weight in downward direction with acceleration a
mg - T = ma
For motion of box on horizontal surface
T = Ma ( M is mass of box )
mg - Ma = ma
a = mg / (m +M)
= 5 x 9.8 / 17
= 2.88 m / s
T = Ma
= 12 x 2.88
= 34.55 N
c )
Horizontal component of the force that the axle exerts on the pulley
= Tension in the wire
= 34,55 N
Vertical component of the force that the axle exerts on the pulley
= Tension in the wire
= 34,55 N
A) The tension in the wire on both sides of the pulley = 32.52 N, 35.45 N
B) The acceleration of the box = 2.71 ms²
C) The horizontal and vertical components of the force = 32.52 N, 56.03 N
Given data :
mass of Block A ( M₁ ) = 12 kg
mass of Block B ( M₂ ) = 5 kg
mass of pulley ( M ) = 2.10 kg
diameter of pulley = 0.560 m
radius of pulley = 0.28 m
B) Calculate the Acceleration of the box
The acceleration of the box can be calculated using this relation below
A = [tex]\frac{m_{2}*g }{[m_{1} + m_{2} + M/2 ]}[/tex]
= ( 5 * 9.8 ) / [ 12 + 5 + 1.05 ]
= ( 49 ) / 18.05
= 2.71 ms⁻²
A) Determine the tension in the wire on both sides of the pulley
Tension in the wire on both sides can be determined by this relation
T₁ = m₁* a
= ( 12 * 2.71 )
= 32.52 N
Also
T₂ ( tension on the other side of the wire )
T₂ = ( m₂ * g ) - m₂* a
= ( 5 * 9.8 ) - ( 5 * 2.71 )
= 35.45 N
C) Determine the horizontal and vertical components of the force
Horizontal component of the force
Fx = T₁ = m₁a
= 32.52 N
Vertical component of the force
Fy = T₂ + Mg
= 35.45 + ( 2.1 * 9.8 )
= 56.03 N
Hence we can conclude that The tension in the wire on both sides of the pulley = 32.52 N, 35.45 N and The acceleration of the box = 2.71 ms², The horizontal and vertical components of the force = 32.52 N, 56.03 N
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