If man is balanced with a system, find Tension, reading of the weighing machine.
Answer
Verified
119.7k+ views
Hint In order to understand the system, split the system into two parts and draw the free body diagram along with the forces. Solve the tension equations and arrive at the reading of the weighing machine if the person pulls the rope.
Complete Step By Step Solution
Let the tension applied on the rope be T. Now, the mass of the man and the mass of the box are the downward force acting on the whole system, assuming that the weight of the machine is negligible. There are two upward forces of T acting on the rope. Draw the FBD of the given scenario.
From the above image you can see the 2 systems, one is the box and the other is the man. Now, let us write the force equations for man system and box system separately.
Let W be the reading measurement of the man inside the box, now
\[T + W - (50 \times g) = 0\](considering Man system)
\[T + W - 500 = 0\](Taking g as \[10m/{s^2}\]) ---------1
Now, let us write the force equations for the second system, that is the box.
\[T - W - (20 \times g) = 0\]
\[T - W - 200 = 0\]---------2
Adding equation 1 and 2 , we cancel out the W term
\[2T - 700 = 0\]
\[T = 350N\]
Now that we have found out the tension of the rope, we can substitute in any of the equations to find the reading of the weighing machine. By substituting in equation 1, we get
\[350 + W - 500 = 0\]
\[W = 150N\]
Note: Weight is given in Newton’s, To convert Newton’s into kilograms , divide by g value
g value is \[10m/{s^2}\]
\[W = 15kg\]
Thus the weight reading on the weighing machine is \[15kg\]
Note
In mechanical terms, tension is defined as the pulling force that are transmitted by a string, cable or wire, etc., to an object axially. Major example of application of tension is the pulley system.
Complete Step By Step Solution
Let the tension applied on the rope be T. Now, the mass of the man and the mass of the box are the downward force acting on the whole system, assuming that the weight of the machine is negligible. There are two upward forces of T acting on the rope. Draw the FBD of the given scenario.
From the above image you can see the 2 systems, one is the box and the other is the man. Now, let us write the force equations for man system and box system separately.
Let W be the reading measurement of the man inside the box, now
\[T + W - (50 \times g) = 0\](considering Man system)
\[T + W - 500 = 0\](Taking g as \[10m/{s^2}\]) ---------1
Now, let us write the force equations for the second system, that is the box.
\[T - W - (20 \times g) = 0\]
\[T - W - 200 = 0\]---------2
Adding equation 1 and 2 , we cancel out the W term
\[2T - 700 = 0\]
\[T = 350N\]
Now that we have found out the tension of the rope, we can substitute in any of the equations to find the reading of the weighing machine. By substituting in equation 1, we get
\[350 + W - 500 = 0\]
\[W = 150N\]
Note: Weight is given in Newton’s, To convert Newton’s into kilograms , divide by g value
g value is \[10m/{s^2}\]
\[W = 15kg\]
Thus the weight reading on the weighing machine is \[15kg\]
Note
In mechanical terms, tension is defined as the pulling force that are transmitted by a string, cable or wire, etc., to an object axially. Major example of application of tension is the pulley system.
Recently Updated Pages
Difference Between Circuit Switching and Packet Switching
Difference Between Mass and Weight
JEE Main Participating Colleges 2024 - A Complete List of Top Colleges
JEE Main Maths Paper Pattern 2025 – Marking, Sections & Tips
Sign up for JEE Main 2025 Live Classes - Vedantu
JEE Main 2025 Exam Pattern: Marking Scheme, Syllabus
Trending doubts
Charging and Discharging of Capacitor
JEE Mains 2025 Correction Window Date (Out) – Check Procedure and Fees Here!
JEE Main 2025 Helpline Numbers for Aspiring Candidates
Electromagnetic Waves Chapter - Physics JEE Main
JEE Main 2023 January 25 Shift 1 Question Paper with Answer Keys & Solutions
JEE Main Maths Class 11 Mock Test for 2025
Other Pages
NCERT Solutions for Class 11 Physics Chapter 7 Gravitation
NCERT Solutions for Class 11 Physics Chapter 4 Laws of Motion
Thermodynamics Class 11 Notes CBSE Physics Chapter 11 (Free PDF Download)
NCERT Solutions for Class 11 Physics Chapter 13 Oscillations
NCERT Solutions for Class 11 Physics Chapter 3 Motion In A Plane
JEE Advanced 2024 Syllabus Weightage