Question

how to find tension when three bodies are in contact

Answer 1

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 Fc 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 T1 acting on the block. Tension T2 acting on the ceiling and Fc the reaction to T2. 

Hence action reaction (Newton's 3 rd law) : |Fc| = |T2| 
We now consider the forces acting on the block (Free Body Diagram) 
Since the block is at rest W + T1 = 0 (Newton's second law, vector equation) 
     W = (0 , -|W|) 
     T1 = (0, |T1|) 
Hence : (0 , -|W|) + (0, |T1|) = 0 
     sum of y coordinates = 0 gives |W| = |T1| 
T1 and T2 represent the tension of the string and their magnitudes are equal. Hence 
    |T2| = |T1| 
T2 and Fc are action and reaction pairs and therefore their magnitudes are equal. Hence 
    |Fc| = |T2| = |T1| = |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 m1 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 stringSolution 
a)




1) free body diagram of block m1 
Newton's second law, assuming m1 accelerating from left to right and |a| is the magnitude of the acceleration. 
     W1 + T1 + N = (m1 |a| , 0) 
with 
     W1 = (0 , - |W1|) = (0 , -m1 g) 
     N = (0 , |N|) 
     T1 = (|T1| , 0) 
above equation in components form: (0 , -m1 g) + (0 , |N|) + (|T1| , 0) = (m1 |a| , 0) 
x components equation 
     0 + 0 + |T1| = m1 |a| 
2) free body diagram of block m2 
Newton's second law, assuming m2 accelerating downward and |a| is the magnitude of the acceleration 
     W2 + T2 = (0 , -m2 |a|) 
     W2 = (0 , - |W2|) = (0 , - m2 g) 
     T2 = (0 , |T2|) 
     (0 , - m2 g) + (0 , |T2|) = (0 , -m2 a) 
y components equation 
     - m2 g + |T2| = - m2 |a| 
Note: |T1| = |T2| tension in the string is the same 
Combining the equations |T1| = m1 |a|, |T1| = |T2| and -m2 g + |T2| = - m2 |a| found above, we can write 
     - m2 g + m1 |a| = - m2 |a| 
Solve for |a| 
     |a| = m2 g / (m1 + m2) 
b) 
Use |T1| = m1 |a| and |a| = m2 g / (m1 + m2) to obtain 
     |T1| = m1 |a| = m1 m2 g / (m1 + m2)