
Several problems with solutions and detailed explanations on systems with strings, pulleys and inclined planes are presented. Free body diagrams of forces, forces expressed by their components and Newton's laws are used to solve these problems. Problems involving forces of friction and tension of strings and ropes are also included.
Problem 1
A block of mass 5 Kg is suspended by a string to a ceiling and is at rest. Find the force F_{c} exerted by the ceiling on the string. Assume the mass of the string to be negligible.
Solution
a)
The
free body diagram below shows the weight W and the tension T_{1} acting on the block. Tension T_{2} acting on the ceiling and F_{c} the reaction to T_{2}.
Hence action reaction (Newton's 3 rd law) : F_{c} = T_{2}
We now consider the forces acting on the block (Free Body Diagram)
Since the block is at rest W + T_{1} = 0 (Newton's second law, vector equation)
W = (0 , W)
T_{1} = (0, T_{1})
Hence : (0 , W) + (0, T_{1}) = 0
sum of y coordinates = 0 gives W = T_{1}
T_{1} and T_{2} represent the tension of the string and their magnitudes are equal. Hence
T_{2} = T_{1}
T_{2} and F_{c} are action and reaction pairs and therefore their magnitudes are equal. Hence
F_{c} = T_{2} = T_{1} = W = m g = 5×10 = 50 N

Problem 2
In the figure below is shown the system below are shown two blocks linked by a string through a pulley, where the block of mass m_{1} slides on the frictionless table. We assume that the string is massless and the pulley is massless and frictionless.
a) Find the magnitude of the acceleration of the two masses
b) Find the tension in the string
Solution
a)
1) free body diagram of block m_{1}
Newton's second law, assuming m_{1} accelerating from left to right and a is the magnitude of the acceleration.
W_{1} + T_{1} + N = (m_{1} a , 0)
with
W1 = (0 ,  W_{1}) = (0 , m_{1} g)
N = (0 , N)
T_{1} = (T_{1} , 0)
above equation in components form: (0 , m_{1} g) + (0 , N) + (T_{1} , 0) = (m_{1} a , 0)
x components equation
0 + 0 + T_{1} = m_{1} a
2) free body diagram of block m_{2}
Newton's second law, assuming m_{2} accelerating downward and a is the magnitude of the acceleration
W_{2} + T_{2} = (0 , m_{2} a)
W_{2} = (0 ,  W_{2}) = (0 ,  m_{2} g)
T_{2} = (0 , T_{2})
(0 ,  m_{2} g) + (0 , T_{2}) = (0 , m_{2} a)
y components equation
 m_{2} g + T_{2} =  m_{2} a
Note: T_{1} = T_{2} tension in the string is the same
Combining the equations T_{1} = m_{1} a, T_{1} = T_{2} and m_{2} g + T_{2} =  m_{2} a found above, we can write
 m_{2} g + m_{1} a =  m_{2} a
Solve for a
a = m_{2} g / (m_{1} + m_{2})
b)
Use T_{1} = m_{1} a and a = m_{2} g / (m1 + m_{2}) to obtain
T_{1} = m_{1} a = m_{1} m_{2} g / (m_{1} + m_{2})

Problem 3
In the two blocks of masses m_{1} and m_{2} and pulley system below, the pulley is frictionless and massless and the string around the pulley is massless. Find an expression of the acceleration when the block are released from rest.
Solution
Let a the magnitude of the acceleration of m_{1} and m_{2} assuming m_{1} accelerating upward and m_{2} accelerating downward.
1) Free body diagram of m_{1}
W_{1} the weight of m_{1}
T_{1} the tension of the string at m_{1} (force exerted by string on m_{1})
a_{1} (vector) acceleration of m_{1}
W_{1} + T_{1} = m_{1} a_{1} ( Newton's second law (vector equation) )
W_{1} = (0 , W_{1})
T_{1} = (0 , T_{1})
a_{1} = (0 , a) acceleration assuming m1 accelerating upward.
(0 , W_{1}) + (0 , T_{1}) = m_{1} (0 , a)
y components equation
W_{1} + T_{1} = m_{1} a equation (1)
2) Free body diagram of m_{2}
W_{2} the weight of m_{2}
T_{2} the tension of the string at m_{2} (force exerted by string on m_{2})
a_{2} (vector) acceleration of m_{2}
W_{2} + T_{2} = m_{2} a_{2} ( Newton's second law (vector equation))
W_{2} = (0 , W_{2})
T_{2} = (0 , T_{2})
a_{2} = (0 ,  a) , assuming m_{2} accelerating downward.
(0 , W_{2}) + (0 , T_{2}) = m_{2} (0 , a)
y components equation
 W_{2} + T_{2} =  m_{2} a equation (2)
Use equation (1) and (2) found above to write
T_{1} = m_{1} a + W_{1}
T_{2} =  m_{2} a + W_{2}
T_{1} and T_{2} represent the tension in the same string and therefore their magnitudes are equal.
Using the fact that T_{1} = T_{2} and the last two equations to write
m_{1} a + W_{1} =  m_{2} a + W_{2}
W_{1} and W_{2} are the weights of m_{1} and m_{2} and are given by
W1 = m_{1} g and W_{2} = m_{2} g
Substitute in the above equation to write
m_{1} a + m_{1} g =  m_{2} a + m_{2} g
Solve for a to obtain
a = g ( m_{2}  m_{1} ) / (m_{1} + m_{2})

Problem 4
Three cords are knotted at point P, with two of these cords fastened to the ceiling making angles α1, α2 and a block of mass m hangs from the third one as shown below.
a) Find the magnitude of the tension in each cord in terms of α1, α2 and m so that the system is at rest.
b) Find numerical values to the three tensions found above for α1 = 45° , α2 = 30° and m = 1 Kg.
Solution
a)
1) Free body diagram of m
W + T '_{3} = 0
W = (0 , W) , weight
T '_{3} = (0 , T '_{3}) , tension of string
y components equation gives
T '_{3} = W
Tension in the same cord
T_{3} = T '_{3} = W
2) Free body diagram of point P
At point p (origin of x y system of axes)
T_{1} + T_{2} + T_{3} = 0 (Newton's second law)
T_{1} = ( T_{1} cos α_{1} , T_{1} sin α_{1})
T_{2} = ( T_{2} cos α_{2} , T_{2} sin α_{2})
T_{3} = (0 , T_{3}) = (0 , W) , (since T_{3} = T '_{3} = W already found above)
sum of x components = 0 gives equation
T_{1} cos α_{1}  T2 cos α_{2} = 0 (equation 1)
sum of y components = 0 gives equation
T_{1} sin α_{1} + T_{2} sin α2  W = 0 (equation 2)
The above is a system of 2 equations with 2 unknown T_{1} and T_{2}.
Solving the system of equations using Cramer's rules, we obtain
T_{1} = W cos α_{2} / ( cos α_{1} sin α_{2} + sin α_{1} cos α_{2} ) = W cos α_{2} / sin(α_{1} + α_{2})
T_{2} = W cos α_{1} / ( cos α_{1} sin α_{2} + sin α_{1} cos α_{2} ) = W cos α_{1} / sin(α_{1} + α_{2})
b)
Numerical application
α1 = 45° , α2 = 30° , m = 1 Kg and g = 10 m/s^{2}
W = 1×10 = 10 N
T_{1} = W cos α_{2} / sin(α_{1} + α_{2}) = 10 cos 30° /sin(45°+30°) = 9.0 N
T_{2} = W cos α_{1} / sin(α_{1} + α_{2}) = 10 cos 45° /sin(45°+30°) = 7.3 N

Problem 5
The system below includes 3 blocks of masses m_{1} = 1 Kg, m2 = 2 Kg and m3 = 5 Kg linked by massless and frictionless strings and pulleys.
a) Find the magnitude of the acceleration of the 3 blocks.
b) Find the magnitude of the tension of the string between m_{1} and m2.
c) Find the magnitude of the tension of the string between m2 and m3.
Solution
a)
We assume that m_{1} is accelerating upward, m_{2} from left to right and m_{3} downward.
We apply Newton's second law for each block. a is the magnitude of the acceleration of all 3 blocks.
1) Free body diagram of m_{1}
W_{1} + T '_{1} = (0 , m_{1} a) (Newton's law)
(0 , w_{1}) + (0 , T '_{1}) = (0 , m_{1} a) (above equation using components, equation 1)
2) Free body diagram of m_{2}
W_{2} + T_{1} + T_{2} + N = (m_{2} a , 0)
(0 ,  W_{2}) + (T_{1} , 0) + (T_{2} , 0) + (0 , N) = (m_{2} a , 0) (above equation using components, equation 2)
3) Free body diagram of m_{3}
W_{3} + T '_{2} = (0 , m_{3} a)
(0 ,  W_{3}) + (0 , T '_{2}) = (0 , m_{3} a) (above equation using components, equation 3)
equation 1 gives: w_{1} + T '_{1} = m_{1} a , y components equal
equation 2 gives: T_{1} + T_{2} = m_{2} a , x components equal
equation 3 gives:  W_{3} +T '_{2} = m_{3} a , y components equal
We also have
T '_{1} = T_{1} same tension in the string between m_{1} and m_{2}
T '_{2} = T_{2} same tension in the string between m_{2} and m_{3}
combining all the equations above we obtain
a = g (m_{3}  m_{1}) / (m_{1} + m_{2} + m_{3})
Use m_{1} = 1 Kg, m_{2} = 2 Kg and m_{3} = 5 Kg and g = 10 m/s^{2}
a = 10 (5  1) / (1 + 2 + 5) = 5 m/s^{2}
b)
T_{1} = T '_{1} = m_{1} a + w_{1}
T_{1} = T '_{1} = 1 × 5 + 1×10 = 15 N
c)
T_{2} = T '_{2} = m_{2} a + T_{1} = m_{2} a + m_{1}a + w_{1}
T_{2} = T '_{2} = 10 + 15 = 25 N

Problem 6
In the system below, blocks of masses m_{1} = 10 Kg and m_{2} = 30 Kg are linked by a massless string through a frictionless pulley.
a) Find the magnitude of the acceleration of the two masses if the coefficient of kinetic friction between the inclined plane and mass m_{1} is equal to 0.4.
b) Find the magnitude of the tension in the string.
Solution
a)
1) Free body diagram of m_{1}
W_{1} + N + F_{k} + T = (m_{1}a , 0) (second law of Newton vector equation) , where a is the mangnitude of the acceleration
where
W_{1} = (W_{1x}, W_{2x}) = (W_{1}sin28° , W_{1}cos28°) , weight of block 1
N = ( 0 , N) , normal force
Fk = (μ_{k} N , 0) , force of friction opposite motion assuming m_{1} moving upward
T = (T , 0) , tension of string.
Rewrite above equation using components
(W_{1}sin28° , W_{1}cos28°) + ( 0 , N) + (μ_{k} N , 0) + (T , 0) = (m_{1}a , 0)
equality of x components
W_{1}sin28° + 0  μ_{k} N + T = m_{1}a , equation (1)
equality of y components
 W_{1}cos28° + N = 0 which gives N = W_{1}cos28° , equation (2)
2) Free body diagram of m_{2}
W_{2} + T ' = (0 ,  m_{2}a) , where a is the magnitude of the acceleration.
where
W_{2} = (0 , W_{2}) , weight of m_{2}
T ' = (0 , T ') , tension of string
rewrite above equation using components
(0 , W_{2}) + (0 , T ') = (0 ,  m_{2}a)
y components equal
 W_{2} + T ' = m_{2}a or T ' = W_{2}  m_{2}a , equation (3)
We now use the fact that the tension in the string is the same in magnitude, hence
T ' = T = W_{2}  m_{2}a
Substitute T by W_{2}  m_{2}a and N by W_{1}cos28° in equation (1) to obtain
 W_{1}sin28°  μ_{k} W_{1}cos28° + W_{2}  m_{2}a = m_{1}a
a (m_{1} + m_{2}) = W_{2} W_{1}sin28°  μ_{k} W_{1}cos28°
a = g (m_{2}  m_{1} sin28°  m_{1} μk cos28°) / (m_{1} + m_{2})
Numerical application with m_{1} = 10 Kg and m_{2} = 30 Kg and μ_{k} = 0.4 , g = 10 m/s^2
a = 10 (30  10(sin28° + 0.4 cos28°)) / (10 + 30) ≈ 5.4 m/s^2
b)
T' = T = W_{2}  m_{2}a = m_{2} g  m_{2} a = 30(10  5.4) ≈ 138 N

