Static

Guide n° 1 of solved exercises of statics

Solve the following exercises

Example, how to calculate forces, power, resistance and weights in pulleys and tackle. Statics problems solved and easy.

problem #1

A body of 200 kgf is lifted by means of a potential tackle with 3 mobile pulleys. What is the value of the power?

Solution

Potential Rigging Problem #1

Statement of exercise n° 1

A body of 200 kgf is lifted by means of a potential tackle with 3 mobile pulleys. What is the value of the power?

Developing

Data:

P = 200 kgf

n = 3

Formulas:

T = P
2n _

Solution

T = 200 kgf
23
T = 200 kgf
8

Result, the power applied to the rig is:

T = 25 kgf

problem #2

A body is supported by a potential 5-pulley tackle. If the applied power is 60 N, what is the weight of the body?

Solution

Potential Rigging Problem #2

Statement of exercise n° 2

A body is supported by a potential 5-pulley tackle. If the applied power is 60 N, what is the weight of the body?

Developing

Data:

T = 60N

n = 5

Formulas:

T = P
2n _

Solution

P = 2n T

P = 2 5 60 N

P = 32 60 N

Result, the weight of the sustained body is:

P = 1920N

Problem #3

Using a 4-pulley factorial rig, a 500-kgf body is balanced. What is the applied power?

Solution

Factorial rigging problem #3

Statement of exercise n° 3

Using a 4-pulley factorial rig, a 500-kgf body is balanced. What is the applied power?

Developing

Data:

P = 500 kgf

n = 4

Formulas:

T = P
2 n

Solution

T = 500 kgf
2 4
T = 500 kgf
8

Result, the power applied to the rig is:

T = 62.5 kgf

Problem #4

Using a lathe with a radius of 12 cm and a crank of 60 cm, a bucket weighing 3.5 kgf, loaded with 12 liters of water, is lifted. What is the applied power?

Solution

Lathe Problem #4

Statement of exercise n° 4

Using a lathe with a radius of 12 cm and a crank of 60 cm, a bucket weighing 3.5 kgf, loaded with 12 liters of water, is lifted. What is the applied power?

Developing

Data:

P: weight of the bucket plus the water.

P = 3.5 kgf + 12 kgf = 15.5 kgf

d1 = 12cm = 0.12m

d2 = 60cm = 0.60m

Formulas:

Equilibrium condition → M F = 0

M F = ∑(F d) = P d 1 + T d 2

Equilibrium condition: The sum of the moments of the forces must be zero: Newton’s first law (equilibrium)

Scheme:

Free body diagram of a lathe
Free body diagram of a lathe

Solution

0 = P d 1 + T d 2

-P d 1 = T d 2

T = -P d 1
d2 _
T = -15.5kgf 0.12m
0.6m

Result, the power applied to the lathe is:

T = -3.1 kgm

T is negative because it rotates clockwise.

Problem #5

In a potential rig with 4 moving pulleys, a force of 30 N is applied to keep the system in equilibrium, it is desired to know the value of the resistance.

Solution

Potential Rigging Problem #5

Statement of exercise n° 5

In a potential rig with 4 moving pulleys, a force of 30 N is applied to keep the system in equilibrium, it is desired to know the value of the resistance.

Developing

Data:

T = 30N

n = 4

Formulas:

T = P
2n _

Solution

T = P
2n _

T 2 n = P

P = 30 N 2 4

Result, the value of the resistance in the rig is:

P = 480N

Problem #6

A body is lifted with a winch with a radius of 30 cm, to which 30 N is applied. What will be the weight of the body if the crank is 90 cm?

Solution

Lathe Problem #6

Statement of exercise n° 6

A body is lifted with a winch with a radius of 30 cm, to which 30 N is applied. What will be the weight of the body if the crank is 90 cm?

Developing

Data:

T = 30N

d1 = 30cm = 0.30m

d2 = 90cm = 0.90m

Formulas:

Equilibrium condition → M F = 0

M F = ∑(F d) = P d 1 + T d 2

Equilibrium condition: The sum of the moments of the forces must be zero: Newton’s first law (equilibrium)

Scheme:

Free body diagram of a lathe

Solution

0 = P d 1 + T d 2

-P d 1 = T d 2

-P = T d 2
d1 _
-P = -30N 0.9m
0.3m

T is negative because it rotates clockwise.

Result, the weight of the lifted body is:

P = 90N

Problem #7

At the ends of a rope, which is on a fixed pulley, two loads of 5 kgf and 7 kgf have been placed. If the radius of the pulley is 12 cm, what is the moment that turns the pulley?

Solution

Fixed Pulley Problem #7

Statement of exercise No. 7

At the ends of a rope, which is on a fixed pulley, two loads of 5 kgf and 7 kgf have been placed. If the radius of the pulley is 12 cm, what is the moment that turns the pulley?

Developing

Data:

F1 = 5 kgf

F2 = 7 kgf

d = 12cm = 0.12m

Formulas:

M F = ∑(F d) = F 1 d 1 + F 2 d 2

Scheme:

Scheme of a fixed pulley

Solution

But d1 = d2 = 0.12m

M F = F 1 d + F 2 d = (F 1 + F 2 ) d

One of the forces must be given a negative direction, normally the one that rotates clockwise, in this case any of them.

M F = (-5kgf + 7kgf) 0.12m

MF = 2 kgf 0.12 m

Result, the moment that makes the pulley rotate is:

MF = 0.24 kgm

Problem #8

Calculate the weight of a body suspended from the rope of a lathe with a radius of 18 cm and a crank 45 cm long, balanced by a force of 60 kgf.

Solution

Lathe Problem #8

Statement of exercise n° 8

Calculate the weight of a body suspended from the rope of a lathe with a radius of 18 cm and a crank 45 cm long, balanced by a force of 60 kgf.

Developing

Data:

T = 60 kgf

d1 = 18cm = 0.18m

d2 = 45cm = 0.45m

Formulas:

Equilibrium condition → M F = 0

M F = ∑(F d) = P d 1 + T d 2

Equilibrium condition: The sum of the moments of the forces must be zero: Newton’s first law (equilibrium)

Scheme:

Free body diagram of a lathe
Free body diagram of a lathe

Solution

0 = P d 1 + T d 2

-P d 1 = T d 2

P = -T d 2
d1 _
P = -(-60kgf) 0.45m
0.18m

T is negative because it rotates clockwise.

Result, the weight of a suspended body is:

P = 150 kgf

Problem #9

What will be the length of the crank of a lathe that, to balance a weight of 150 kgf, it is necessary to apply a force of 40 kgf? The radius of the cylinder is 20 cm.

Solution

Lathe Problem #9

Statement of exercise n° 9

What will be the length of the crank of a lathe that, to balance a weight of 150 kgf, it is necessary to apply a force of 40 kgf? The radius of the cylinder is 20 cm.

Developing

Data:

P = 150 kgf

T = 40 kgf

d1 = 20cm = 0.20m

Formulas:

Equilibrium condition → M F = 0

M F = ∑(F d) = P d 1 + T d 2

Equilibrium condition: The sum of the moments of the forces must be zero: Newton’s first law (equilibrium)

Scheme:

Free body diagram of a lathe
Free body diagram of a lathe

Solution

0 = P d 1 + T d 2

-P d 1 = T d 2

d2 = _ -P d 1
T

T is negative because it rotates clockwise.

d2 = _ -150kgf 0.20m
-40kgf

Result, the length of the lathe handle should be:

d2 = 0.75m

Problem #10

A body is lifted with a winch of radius 20 cm, to which 40 kgf is applied. What will be the weight of the body if the crank is 80 cm?

Solution

Lathe Problem #10

Statement of exercise n° 10

A body is lifted with a winch of radius 20 cm, to which 40 kgf is applied. What will be the weight of the body if the crank is 80 cm?

Developing

Data:

T = 40 kgf

d1 = 20cm = 0.20m

d2 = 80cm = 0.80m

Formulas:

Equilibrium condition → M F = 0

M F = ∑(F d) = P d 1 + T d 2

Equilibrium condition: The sum of the moments of the forces must be zero: Newton’s first law (equilibrium)

Scheme:

Free body diagram of a lathe

Solution

0 = P d 1 + T d 2

-P d 1 = T d 2

-P = T d 2
d1 _
-P = -40kgf 0.8m
0.2m

T is negative because it rotates clockwise.

Result, the weight of the lifted body is:

W = 160 kgm

 

Related Articles

Leave a Reply

Your email address will not be published. Required fields are marked *

Check Also
Close
Back to top button