Increased acceleration in a pully system

In summary: The equation in red is irrelevant here. The person exerts F force, and the tension in the chord equals to Ft=F. The person in the bucket is pulled by 2Ft upward, and gravity pulls downward. So the equation you wrote 2Ft - mg = ma is correct. The other equation refers to the case of a pulley with two masses.
  • #1
coneheadceo
25
0

Homework Statement


A window washer pulls themself upward using a bucket/single pulley system. Their weight with the bucket is 65 Kg.
a) How hard must she pull doward to raise herself slowly at a constant speed?
b) If they increase the force by 15% what will her accelertion be?

Homework Equations


2Ft - mg = ma

a = (Mb - Mp)/(Mb+Mp) ... I think

The Attempt at a Solution


a) I've got no problem...

to get constant speed accel = 0 so...

2Ft - mg = 0
2Ft = mg
Ft = mg/2 -> (65)(9.8)/2 = 320 N

b) This is where I think I have the wrong equation to attempt this as
(320)(0.15)= 48, so ... [(368-320)/(368+320)](9.8)= 0.68 m/s^2

answer in the book is 1.5 m/s^2 so need something else but don't know what to use.
 
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  • #2
coneheadceo said:

Homework Statement


A window washer pulls themself upward using a bucket/single pulley system. Their weight with the bucket is 65 Kg.
a) How hard must she pull doward to raise herself slowly at a constant speed?
b) If they increase the force by 15% what will her accelertion be?

Homework Equations


2Ft - mg = ma

a = (Mb - Mp)/(Mb+Mp) ... I think

The Attempt at a Solution


a) I've got no problem...

to get constant speed accel = 0 so...

2Ft - mg = 0
2Ft = mg
Ft = mg/2 -> (65)(9.8)/2 = 320 N

b) This is where I think I have the wrong equation to attempt this as
(320)(0.15)= 48, so ... [(368-320)/(368+320)](9.8)= 0.68 m/s^2

answer in the book is 1.5 m/s^2 so need something else but don't know what to use.

The equation in red is irrelevant here. The person exerts F force, and the tension in the chord equals to Ft=F. The person in the bucket is pulled by 2Ft upward, and gravity pulls downward. So the equation you wrote 2Ft - mg = ma is correct.

The other equation refers to the case of a pulley with two masses.

ehild
 
  • #3
Your new upward force is 2(320+48)
 
  • #4
So am I literally taking (9.8 m/s^2)(0.15)= 1.47 m/s^2 to get the acceleration?
 
  • #5
coneheadceo said:
So am I literally taking (9.8 m/s^2)(0.15)= 1.47 m/s^2 to get the acceleration?

Yes.

ehild
 

1. What factors affect the acceleration in a pulley system?

The acceleration in a pulley system is affected by the mass of the object being moved, the tension in the rope, and the friction between the rope and the pulley.

2. How does increasing the number of pulleys affect the acceleration in a pulley system?

Increasing the number of pulleys will decrease the acceleration in a pulley system. This is because with each additional pulley, more of the force is used to support the weight of the pulleys themselves instead of accelerating the object.

3. Can the acceleration in a pulley system be greater than the acceleration due to gravity?

No, the acceleration in a pulley system cannot be greater than the acceleration due to gravity. The acceleration due to gravity is a constant, while the acceleration in a pulley system is affected by various factors such as mass and tension.

4. How does the angle of the rope affect the acceleration in a pulley system?

The angle of the rope can affect the acceleration in a pulley system. As the angle of the rope decreases, the tension in the rope increases, resulting in a greater acceleration. However, if the angle becomes too small, it can cause the rope to slip and decrease the acceleration.

5. What is the relationship between the acceleration and velocity in a pulley system?

The acceleration and velocity in a pulley system have an inverse relationship. This means that as the acceleration increases, the velocity decreases, and vice versa. This is due to the conservation of energy in the system.

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