Question
Download Solution PDFThe ________ is equal to the product of the force applied and radius of the shaft.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
Torque
- It is the turning moment of a wheel measured in N-m and is the power required to overcome the resistance to move.
- Torque is a product of the force applied at a tangent and the radius of the shaft.
Additional Information
Moment of a force
Varingon’s theorem states that if many coplanar forces are acting on a body, then the algebraic sum of moments of all the forces about a point in the plane of the forces is equal to the moment of their resultant about the same point.
- The moment of the resultant force about point O is given as,
- Mo = (F1a) + (F2b) + (F3c) + M1 + M2
- The algebraic sum of moments of the forces forming a couple about any point in their plane is equal to the moment of the couple.
- The moment of a force about any point is the Product of force and the perpendicular distance between the point and line of action of the force.
Last updated on May 19, 2025
-> JKSSB Junior Engineer recruitment exam date 2025 for Civil and Electrical Engineering has been released on its official website.
-> JKSSB JE exam will be conducted on 10th August (Civil), and on 27th July 2025 (Electrical).
-> JKSSB JE recruitment 2025 notification has been released for Civil Engineering.
-> A total of 508 vacancies has been announced for JKSSB JE Civil Engineering recruitment 2025.
-> JKSSB JE Online Application form will be activated from18th May 2025 to 16th June 2025
-> Candidates who are preparing for JKSSB JE can download syllabus PDF from official website of JKSSB.
-> The candidates can check the JKSSB JE Previous Year Papers to understand the difficulty level of the exam.
-> Candidates also attempt the JKSSB JE Mock Test which gives you an experience of the actual exam.