Answer:
Tension, T1 = 46.2 N and T2 = 52 N, where as acceleration = 3.3 ms^-2.
Forces on the pulley are 46.2 N , 81.4 N horizontal and vertical
respectively.
Step-by-step explanation:
Given:
Mass of the box on rest,
= 14 kg
Mass of the attached box,
= 8 kg
Mass of the pulley,
= 3 kg
Diameter of the pulley,
= 1 m
Radius of the pulley,
= 0.5 m
Here we will be using the concept of net force (
),net torque (
) and acceleration of the pulley .
A FBD is attached with.
Lets find the tension on the wire using Fnet.
⇒
...for m1
⇒
can be written as
...for m2
Considering clockwise torque as negative and anticlockwise torque as positive.
Moment of inertia (I) of the disk/pulley =
and
.
Now using net torque on the pulley we can say that.
⇒

⇒
⇒

⇒ Plugging T1 and T2 .
⇒

⇒ Isolating a from the rest.
⇒

⇒
![m_2g=a\ [(m_3)/(2)+m_2+m_1]](https://img.qammunity.org/2021/formulas/physics/high-school/z4c6ewib8pwk7zxwwozapwcogxmo576xsl.png)
⇒

⇒ Plugging the numeric value
⇒

⇒

⇒ Acceleration = 3.3

So,
(a).
Tension in the wire
⇒

⇒

(b).
The acceleration of the box is 3.3 ms^-2.
(c).
Forces on the pulley.
Horizontal force,
=
=

Vertical force,
=
=
=
The values are as follows:
Tension as T1 = 46.2 N and T2 = 52 N ,where as acceleration =3.3 ms^-2.
Forces on the pulley are 46.2 N , 81.4 N horizontal and vertical
respectively.