Join Sarthaks eConnect Today - Largest Online Education Community!
0 votes
asked in Physics by (79.2k points)

Two opposite forces F1 = 120N and F2 = 80N act on an heavy elastic plank of modulus of elasticity y = 2×1011 N/m2 and length L = 1m placed over a smooth horizontal surface. The cross-sectional area of plank is A = 0.5m2 . If the change in the length of plank is x × 10–9m, then find x ?

1 Answer

+1 vote
answered by (70.9k points)
selected by
Best answer

T = F1 – (F1 – F2 ) x/L

Happy Diwali Wishes

Related questions

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.

One Thought Forever

“There is a close connection between getting up in the world and getting up in the morning.“
– Anon