Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
2.1k views
in Physics by (49.4k points)

Five bodies 1 kg wt., 2kg wt., 3kg wt., 4kg wt. and 5kg wt. are suspended at a distances of 1m, 2m, 3m, 4m, and 5 m from one end of a uniform rod of mass 4kg and 6 metres length. Find the point about which the rod will balance about the knife edge placed at a distance of ‘x’ from the end A.

1 Answer

0 votes
by (53.6k points)
selected by
 
Best answer

Resultant force acting vertically downwards on the rod

=P + Q + R + S + T + W

= 1 + 2 + 3 + 4 + 5 + 4 = 19 kg wt

Thus force of 19 kgwt. acting vertically upwards will balance the rod. This force acts at a distance ‘x’ from A.

i.e., R1 = 19 kgwt.

Taking the moments about A,

P × AB + Q × AC + R × AD + S × AE + T × AF + W × Al = R1x.

1 × 1 + 2 × 2 + 3 × 3 + 4 × 4 + 5 × 5 + 4 × 3 = R1x = 19x.

1 + 4 + 9 + 16 + 25 + 12 = 19x.

67 = 19x or x = \(\frac {67}{19}\) = 3.52m

∴ Distance of the knife edge from the end A is 3.52m.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...