Class 11 Exam > Class 11 Questions > A weightless rod is acted on by upward parall...
Start Learning for Free
A weightless rod is acted on by upward parallel forces of 2N and 4N ends A and B respectively. The total length of the rod AB = 3m. To keep the rod in equilibrium a force of 6N should act in the following manner :
- a)Downwards at any point between A and B.
- b)Downwards at mid point of AB.
- c)Downwards at a point C such that AC = 1m.
- d)Downwards at a point D such that BD = 1m.
Correct answer is option 'D'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
A weightless rod is acted on by upward parallel forces of 2N and 4N en...
Most Upvoted Answer
A weightless rod is acted on by upward parallel forces of 2N and 4N en...
Free Test
FREE
| Start Free Test |
Community Answer
A weightless rod is acted on by upward parallel forces of 2N and 4N en...
Given:
- Length of rod AB = 3m
- Upward forces acting on the rod: 2N at end A and 4N at end B
- A force of 6N is required to keep the rod in equilibrium
To keep the rod in equilibrium, the net force acting on the rod should be zero.
Solution:
We can find the position where the force of 6N should act by taking moments about different points on the rod.
Taking moments about end A:
- Moment due to force of 2N at A = 2N x 0m = 0
- Moment due to force of 4N at B = 4N x 3m = 12Nm
- Moment due to force of 6N at point X = 6N x AX
- Net moment = 0 + 12Nm + 6N x AX = 0
- AX = -2m (which is not possible as it lies outside the rod)
Taking moments about end B:
- Moment due to force of 4N at B = 4N x 0m = 0
- Moment due to force of 2N at A = 2N x 3m = 6Nm
- Moment due to force of 6N at point Y = 6N x BY
- Net moment = 0 + 6Nm + 6N x BY = 0
- BY = 1m
Taking moments about mid-point of the rod:
- Moment due to force of 2N at A = 2N x 1.5m = 3Nm
- Moment due to force of 4N at B = 4N x 1.5m = 6Nm
- Moment due to force of 6N at point Z = 6N x CZ
- Net moment = 3Nm + 6Nm + 6N x CZ = 0
- CZ = -1.5m (which is not possible as it lies outside the rod)
Taking moments about a point C such that AC = 1m:
- Moment due to force of 2N at A = 2N x 1m = 2Nm
- Moment due to force of 4N at B = 4N x 2m = 8Nm
- Moment due to force of 6N at point C = 6N x 0m = 0
- Net moment = 2Nm + 8Nm + 0 = 0
- Therefore, the force of 6N should act downwards at point C.
Taking moments about a point D such that BD = 1m:
- Moment due to force of 2N at A = 2N x 2m = 4Nm
- Moment due to force of 4N at B = 4N x 0m = 0
- Moment due to force of 6N at point D = 6N x BD = 6N x 1m = 6Nm
- Net moment = 4Nm + 0 + 6Nm = 0
- Therefore, the force of 6N should act downwards at point D.
Conclusion:
To keep the rod in equilibrium, the force of 6N should act downwards at a point D such that BD = 1
- Length of rod AB = 3m
- Upward forces acting on the rod: 2N at end A and 4N at end B
- A force of 6N is required to keep the rod in equilibrium
To keep the rod in equilibrium, the net force acting on the rod should be zero.
Solution:
We can find the position where the force of 6N should act by taking moments about different points on the rod.
Taking moments about end A:
- Moment due to force of 2N at A = 2N x 0m = 0
- Moment due to force of 4N at B = 4N x 3m = 12Nm
- Moment due to force of 6N at point X = 6N x AX
- Net moment = 0 + 12Nm + 6N x AX = 0
- AX = -2m (which is not possible as it lies outside the rod)
Taking moments about end B:
- Moment due to force of 4N at B = 4N x 0m = 0
- Moment due to force of 2N at A = 2N x 3m = 6Nm
- Moment due to force of 6N at point Y = 6N x BY
- Net moment = 0 + 6Nm + 6N x BY = 0
- BY = 1m
Taking moments about mid-point of the rod:
- Moment due to force of 2N at A = 2N x 1.5m = 3Nm
- Moment due to force of 4N at B = 4N x 1.5m = 6Nm
- Moment due to force of 6N at point Z = 6N x CZ
- Net moment = 3Nm + 6Nm + 6N x CZ = 0
- CZ = -1.5m (which is not possible as it lies outside the rod)
Taking moments about a point C such that AC = 1m:
- Moment due to force of 2N at A = 2N x 1m = 2Nm
- Moment due to force of 4N at B = 4N x 2m = 8Nm
- Moment due to force of 6N at point C = 6N x 0m = 0
- Net moment = 2Nm + 8Nm + 0 = 0
- Therefore, the force of 6N should act downwards at point C.
Taking moments about a point D such that BD = 1m:
- Moment due to force of 2N at A = 2N x 2m = 4Nm
- Moment due to force of 4N at B = 4N x 0m = 0
- Moment due to force of 6N at point D = 6N x BD = 6N x 1m = 6Nm
- Net moment = 4Nm + 0 + 6Nm = 0
- Therefore, the force of 6N should act downwards at point D.
Conclusion:
To keep the rod in equilibrium, the force of 6N should act downwards at a point D such that BD = 1
Attention Class 11 Students!
To make sure you are not studying endlessly, EduRev has designed Class 11 study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Class 11.
|
Explore Courses for Class 11 exam
|
|
Similar Class 11 Doubts
Top Courses for Class 11View all
A weightless rod is acted on by upward parallel forces of 2N and 4N ends A and B respectively. The total length of the rod AB = 3m. To keep the rod in equilibrium a force of 6N should act in the following manner :a)Downwards at any point between A and B.b)Downwards at mid point of AB.c)Downwards at a point C such that AC = 1m.d)Downwards at a point D such that BD = 1m.Correct answer is option 'D'. Can you explain this answer?
Question Description
A weightless rod is acted on by upward parallel forces of 2N and 4N ends A and B respectively. The total length of the rod AB = 3m. To keep the rod in equilibrium a force of 6N should act in the following manner :a)Downwards at any point between A and B.b)Downwards at mid point of AB.c)Downwards at a point C such that AC = 1m.d)Downwards at a point D such that BD = 1m.Correct answer is option 'D'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A weightless rod is acted on by upward parallel forces of 2N and 4N ends A and B respectively. The total length of the rod AB = 3m. To keep the rod in equilibrium a force of 6N should act in the following manner :a)Downwards at any point between A and B.b)Downwards at mid point of AB.c)Downwards at a point C such that AC = 1m.d)Downwards at a point D such that BD = 1m.Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A weightless rod is acted on by upward parallel forces of 2N and 4N ends A and B respectively. The total length of the rod AB = 3m. To keep the rod in equilibrium a force of 6N should act in the following manner :a)Downwards at any point between A and B.b)Downwards at mid point of AB.c)Downwards at a point C such that AC = 1m.d)Downwards at a point D such that BD = 1m.Correct answer is option 'D'. Can you explain this answer?.
A weightless rod is acted on by upward parallel forces of 2N and 4N ends A and B respectively. The total length of the rod AB = 3m. To keep the rod in equilibrium a force of 6N should act in the following manner :a)Downwards at any point between A and B.b)Downwards at mid point of AB.c)Downwards at a point C such that AC = 1m.d)Downwards at a point D such that BD = 1m.Correct answer is option 'D'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A weightless rod is acted on by upward parallel forces of 2N and 4N ends A and B respectively. The total length of the rod AB = 3m. To keep the rod in equilibrium a force of 6N should act in the following manner :a)Downwards at any point between A and B.b)Downwards at mid point of AB.c)Downwards at a point C such that AC = 1m.d)Downwards at a point D such that BD = 1m.Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A weightless rod is acted on by upward parallel forces of 2N and 4N ends A and B respectively. The total length of the rod AB = 3m. To keep the rod in equilibrium a force of 6N should act in the following manner :a)Downwards at any point between A and B.b)Downwards at mid point of AB.c)Downwards at a point C such that AC = 1m.d)Downwards at a point D such that BD = 1m.Correct answer is option 'D'. Can you explain this answer?.
Solutions for A weightless rod is acted on by upward parallel forces of 2N and 4N ends A and B respectively. The total length of the rod AB = 3m. To keep the rod in equilibrium a force of 6N should act in the following manner :a)Downwards at any point between A and B.b)Downwards at mid point of AB.c)Downwards at a point C such that AC = 1m.d)Downwards at a point D such that BD = 1m.Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11.
Download more important topics, notes, lectures and mock test series for Class 11 Exam by signing up for free.
Here you can find the meaning of A weightless rod is acted on by upward parallel forces of 2N and 4N ends A and B respectively. The total length of the rod AB = 3m. To keep the rod in equilibrium a force of 6N should act in the following manner :a)Downwards at any point between A and B.b)Downwards at mid point of AB.c)Downwards at a point C such that AC = 1m.d)Downwards at a point D such that BD = 1m.Correct answer is option 'D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A weightless rod is acted on by upward parallel forces of 2N and 4N ends A and B respectively. The total length of the rod AB = 3m. To keep the rod in equilibrium a force of 6N should act in the following manner :a)Downwards at any point between A and B.b)Downwards at mid point of AB.c)Downwards at a point C such that AC = 1m.d)Downwards at a point D such that BD = 1m.Correct answer is option 'D'. Can you explain this answer?, a detailed solution for A weightless rod is acted on by upward parallel forces of 2N and 4N ends A and B respectively. The total length of the rod AB = 3m. To keep the rod in equilibrium a force of 6N should act in the following manner :a)Downwards at any point between A and B.b)Downwards at mid point of AB.c)Downwards at a point C such that AC = 1m.d)Downwards at a point D such that BD = 1m.Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of A weightless rod is acted on by upward parallel forces of 2N and 4N ends A and B respectively. The total length of the rod AB = 3m. To keep the rod in equilibrium a force of 6N should act in the following manner :a)Downwards at any point between A and B.b)Downwards at mid point of AB.c)Downwards at a point C such that AC = 1m.d)Downwards at a point D such that BD = 1m.Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A weightless rod is acted on by upward parallel forces of 2N and 4N ends A and B respectively. The total length of the rod AB = 3m. To keep the rod in equilibrium a force of 6N should act in the following manner :a)Downwards at any point between A and B.b)Downwards at mid point of AB.c)Downwards at a point C such that AC = 1m.d)Downwards at a point D such that BD = 1m.Correct answer is option 'D'. Can you explain this answer? tests, examples and also practice Class 11 tests.
|
|
Explore Courses for Class 11 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup