Guide n° 2 of exercises of statics
Solve the following exercises
problem #1
Two 10-N weights are suspended at the ends of a rope that passes over a light, frictionless pulley. The pulley is attached to a chain that hangs from the ceiling. Decide:
a) The tension in the string.
b) The chain tension.
Solution to problem #1
Statement of exercise n° 1
Two 10-N weights are suspended at the ends of a rope that passes over a light, frictionless pulley. The pulley is attached to a chain that hangs from the ceiling. Decide:
a) The tension in the string.
b) The chain tension.
Developing
Data:
P1 = P2 = 10N
Formulas:
P = m g
∑F = 0 (equilibrium condition)
Equilibrium condition: The sum of the moments of the forces must be zero: Newton’s first law (equilibrium)
Solution
First we make a scheme to correctly visualize the statement:
to)
We make the free body diagram for one of the strings:
We set the equilibrium conditions:
∑F = 0
P1 – T1 = 0
We clear T 1 :
T1 = P1 _
We replace by the value of P:
T1 = 10N
The same result will be obtained for T 2 :
Result, string tension is:
T1 = T2 = 10N
b)
We draw the free body diagram for the chain:
We set the equilibrium conditions:
∑F = 0
P – T = 0
In this case the weight P is the sum of the weights hanging from the strings:
P = P1 + P2
P = 10N + 10N
P = 20N
We clear the tension T:
T = P
Result, chain tension is:
T = 20N
problem #2
In a mechanical workshop, the engine of a car, whose weight is 350 kgf, is lifted by means of a differential hoist. If the radii of the pulleys are R = 15 cm and r = 12 cm, what is the force that balances that weight?
Solution to problem #2
Statement of exercise n° 2
In a mechanical workshop, the engine of a car, whose weight is 350 kgf, is lifted by means of a differential hoist. If the radii of the pulleys are R = 15 cm and r = 12 cm, what is the force that balances that weight?
Developing
Data:
P = 350 kgf = 3,432.3275 N
R=15cm=0.15m
r = 12cm = 0.12m
Formulas:
| F = | R – r | ·P |
| 2 R |
Solution
We directly apply the differential rigging formula, replacing it with the data after converting the units:
| F = | 0.15m-0.12m | 3,432.3275N |
| 2 0.15m |
| F = | 0.03m 3,432.3275N |
| 0.3m |
F = 343.23275N
Result, the force that balances that weight of the motor is:
F = 343.2N
Problem #3
The radii of a differential rig are R = 20 cm and r = 15 cm. If a force of 80 kgf is applied, what is the weight of the body that balances it?
Solution to problem #3
Statement of exercise n° 3
The radii of a differential rig are R = 20 cm and r = 15 cm. If a force of 80 kgf is applied, what is the weight of the body that balances it?
Developing
Data:
F = 80 kgf = 784.532 N
R=20cm=0.2m
r = 15cm = 0.15m
Formulas:
| F = | R – r | ·P |
| 2 R |
Scheme:
Solution
We clear the weight of the differential rigging formula:
| P = | F 2 R |
| R – r |
We replace by the data after converting the units:
| P = | 313.8128 N m |
| 0.05m |
P = 6,276.256N
Result, the body weight is:
P = 6,276.3N
Problem #4
In a 2nd gender lever, it has a power arm Bp = 1.20 m and a resistance arm Br = 8 cm. Then the multiplication of the lever is:
a) 6
b) 12
c) 15
c) 20
Solution to problem #4
Statement of exercise n° 4
In a 2nd gender lever, it has a power arm Bp = 1.20 m and a resistance arm Br = 8 cm. Then the multiplication of the lever is:
a) 6
b) 12
c) 15
c) 20
Developing
Data:
Bp = 1.20m
Br = 8cm = 0.08m
Formulas:
| M = | bp |
| br |
Solution
Without calculating the moments, the multiplication ratio is always the power arm over the resistance arm, so:
| M = | 1.20m |
| 0.08m |
Result, the multiplication of the lever is:
M = 15
Problem #5
Calculate at what distance from a 60 kgf power a rigid iron bar will be supported, to balance a 300 kgf box that is 0.75 m from the support.
Solution to problem #5
Statement of exercise n° 5
Calculate at what distance from a 60 kgf power a rigid iron bar will be supported, to balance a 300 kgf box that is 0.75 m from the support.
Developing
Data:
F = 60 kgf = 588.399 N
P = 300 kgf = 2,941.995 N
BP = 0.75m
Formulas:
MF = 0 (Equilibrium condition of moments)
F B F = P B P
Solution
It is understood that it is a 1st class lever.
From the lever equation we isolate the arm of the force (B F ):
| B F = | P B P |
| F |
We substitute for the data and calculate:
| B F = | 2,941.995N 0.75m |
| 588,399N |
| B F = | 2,206.49625m |
| 588,399 |
Result, the distance between the stem and the support point is:
BF = 3.75m
Problem #6
Calculate the power that is necessary to apply to a fixed pulley, to lift a weight of 80 kgf.
Solution to problem #6
Statement of exercise n° 6
Calculate the power that is necessary to apply to a fixed pulley, to lift a weight of 80 kgf.
Developing
Data:
P = 80 kgf = 784.5 N
Formulas:
∑F = 0 (equilibrium condition)
Equilibrium condition: The sum of the moments of the forces must be zero: Newton’s first law (equilibrium)
Solution
First we make a scheme to correctly visualize the statement:
We apply the equilibrium condition taking into account that in the fixed pulley the weight and the power are equal and opposite forces:
∑F = 0
P – Q = 0
Q=P
Result, the power applied to the pulley is:
Q = 784.5N
Problem #7
What power will be applied to balance a resistance of 90 kgf, by means of a mobile pulley?
Solution to problem #7
Statement of exercise No. 7
What power will be applied to balance a resistance of 90 kgf, by means of a mobile pulley?
Developing
Data:
P = 90 kgf = 882.6 N
Formulas:
Q = ½ P
Solution
We make the scheme:
We apply the simple mobile pulley formula:
Q = ½ P
We substitute and calculate:
Q = ½ 882.6 N
Result, the power applied to the pulley is:
Q = 441.3N
Problem #8
A rope is wrapped around a cylinder with a diameter of 30 cm (which can rotate around an axis). If a force (weight) of 1.8 kgf is applied to it, what is the value of the moment that rotates the cylinder?
Statement of exercise n° 8
A rope is wrapped around a cylinder with a diameter of 30 cm (which can rotate around an axis). If a force (weight) of 1.8 kgf is applied to it, what is the value of the moment that rotates the cylinder?
Solution to problem #8
Developing
Data:
P = 1.8 kgf = 17.65197 N
d = 30cm = 0.3m
Formulas:
M P = P r
Solution
We know that:
d = 2 r
r = ½ d
r = ½ 0.3m
r = 0.15m
We apply the formula for now:
M P = P r
We substitute for the data and calculate:
M P = 17.65197 N 0.15 m
M P = 2.6477955 N m
Result, the moment produced by the weight that rotates the cylinder is:
M P = 2.65 N m