# Upward Acceleration of Hanging Mass Connected Pulley

• MissBisson
In summary, a 27.1 kg block is connected to a 6.70 kg block by a massless string, with a force of 213.7 N acting on the first block at an angle of 34.5°. The frictionless pulley has a radius of 0.097 m and a moment of inertia of 0.070 kgm2. Using the equations for torque and acceleration, it is determined that the upward acceleration of the second block is 2.68 m/s^2. The pulley must accelerate counter clockwise in order for this to be true.
MissBisson

## Homework Statement

A 27.1 kg block (m1) is on a horizontal surface, connected to a 6.70 kg block (m2) by a massless string as shown in the figure below. The frictionless pulley has a radius R = 0.097 m and a moment of inertia I=0.070 kgm2. A force F = 213.7 N acts on m1 at an angle theta = 34.5°. There is no friction between m1 and the surface. What is the upward acceleration of m2?

## Homework Equations

Torque = F x R
Torque = I x alpha
a = R x alpha
I = 1/2 m x R^2

## The Attempt at a Solution

Block 1 : Fcos(theta) - T1 = m1a T1 = Fcos(theta) - m1a (opposite direction of accereration)
Block 2 : T2 - m2g = m2a T2 = m2a + m2g (in the direction of accleration)

Torque = (T2-T1)R = I x (a/R)
(m2a + m2g - Fcos(theta) + m1a)R = 1/2 m x R^2 x (a/R)
m2a + m2g - Fcos(theta) + m1a = (m x R^2 x a)/(2 x R^2) R^2 cancel out
m2a + m2g - Fcos(theta) + m1a = (m x a)/2
m2a + m1a - (m x a)/2 = Fcos(theta) - m2g
a (m2 + m1 - m/2) = Fcos(theta) - m2g
a = (Fcos(theta) - m2g)/(m2 + m1 - m/2) -----------------> A
a = (213.7cos(34.5) - 6.7*9.81)/ (6.7 + 27.1 - 17.88/2)
a = (110.39/26.36)
a= 4.187m/s^2

I know that the answer is 2.68m/s^2 and that the equation A should be

a = (Fcos(theta) - m2g)/(m2 + m1 + m/2)

But i don't know how to get that...

#### Attachments

• prob24_2masspulley.gif
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MissBisson said:

## Homework Statement

A 27.1 kg block (m1) is on a horizontal surface, connected to a 6.70 kg block (m2) by a massless string as shown in the figure below. The frictionless pulley has a radius R = 0.097 m and a moment of inertia I=0.070 kgm2. A force F = 213.7 N acts on m1 at an angle theta = 34.5°. There is no friction between m1 and the surface. What is the upward acceleration of m2?

## Homework Equations

Torque = F x R
Torque = I x alpha
a = R x alpha
I = 1/2 m x R^2

## The Attempt at a Solution

Block 1 : Fcos(theta) - T1 = m1a T1 = Fcos(theta) - m1a (opposite direction of accereration)
Block 2 : T2 - m2g = m2a T2 = m2a + m2g (in the direction of accleration)

Torque = (T2-T1)R = I x (a/R)
If m2 is supposed to accelerate upwards then m1 must be accelerating leftwards, right? That means the pulley should be accelerating counter clockwise. That would make T1 > T2 if you want to maintain the assumption of m2 accelerating upwards. So, rearrange your T1 and T2 in the above and redo from there.
I know that the answer is 2.68m/s^2 and that the equation A should be
Strange, that's not the value that I get. Let's see how you do and then we can compare notes :)

gneill said:
If m2 is supposed to accelerate upwards then m1 must be accelerating leftwards, right? That means the pulley should be accelerating counter clockwise. That would make T1 > T2 if you want to maintain the assumption of m2 accelerating upwards. So, rearrange your T1 and T2 in the above and redo from there.

Strange, that's not the value that I get. Let's see how you do and then we can compare notes :)

When I did

Torque = (T1-T2)R = I x (a/R)

It worked out perfectly :) thank you

## What is upward acceleration of hanging mass connected pulley?

The upward acceleration of hanging mass connected pulley refers to the rate at which the hanging mass connected to a pulley system increases its velocity in the upward direction. This acceleration is caused by the force of gravity acting on the hanging mass and the tension force in the rope or cable connected to the pulley.

## How is upward acceleration of hanging mass connected pulley calculated?

The upward acceleration of hanging mass connected pulley can be calculated using the formula a = (m1g - m2g)/ (m1 + m2), where m1 is the mass of the hanging object, m2 is the mass of the pulley, and g is the acceleration due to gravity (9.8 m/s²).

## What factors affect the upward acceleration of hanging mass connected pulley?

The upward acceleration of hanging mass connected pulley is affected by the mass of the hanging object and the pulley, the force of gravity, and the tension in the rope or cable connected to the pulley. Other factors that may influence the acceleration include friction in the pulley system and air resistance.

## How does the angle of the rope or cable affect the upward acceleration of hanging mass connected pulley?

The angle of the rope or cable can affect the upward acceleration of hanging mass connected pulley. As the angle increases, the tension in the rope or cable decreases, leading to a lower acceleration. This is because a greater angle results in a smaller component of the tension force acting in the upward direction.

## What are the real-life applications of understanding upward acceleration of hanging mass connected pulley?

Understanding upward acceleration of hanging mass connected pulley has various real-life applications, such as in elevators, cranes, and construction equipment. It is also essential in understanding the mechanics of sports activities like rock climbing and weightlifting. Additionally, this concept is crucial in designing and optimizing pulley systems for various tasks in industries such as manufacturing and transportation.

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