Find worked solutions for different exam papers, ask questions and connect with
medical aspirants and students from all over the world.

Home Forums IMAT Exam IMAT – Problem Solving and Discussion IMAT 2013 Q54 Physics and Mathematics

  • IMAT 2013 Q54 Physics and Mathematics

  • Nicolas Sanchez

    June 21, 2021 at 9:44 am
    968 Replies
    159150 XP
    Contributions: EnterMedSchool Donor Team Manager Volunteer
    Reputation: Level 7

    A uniform beam, 3.0m long, of weight 100N has a 300N weight placed 0.50m from one end. The beam is suspended by a string 1.0m from the same end.
    A diagram of the weights placed on the beam is given below:

    [See diagram below]

    How far from the other end must a weight of 80N be placed for the beam to be balanced?

    A. 0.75m

    B. 2.25m

    C. 1.25m

    D. 1.875m

    E. 0.125m


    July 15, 2021 at 10:51 pm
    984 Replies
    59071 XP
    Contributions: Volunteer
    Reputation: Level 6
    Down Accepted answer

    Answer= A

    You have to use concept of center of mass. You can also solve it by considering movement of beam like clockwise and anti clockwise.

    Let us consider x as distance for 80N from other end to be balanced.

    Concept of center of mass stats that-

    m1x1 + m2x2= m3x3

    Let m1= 100N and x1= 0.5m.

    The 100N weight will be balanced by string with length (1-0.5)m.

    Let m2= 80 N and x2= (2-x)

    The 80N weight will be balanced by (2-x)m because its placed at that end.

    The 300N weight will be balanced by string with length (1-0.5)m because its placed on that end.

    So, Let m3= 300N and x3= 0.5m

Sort by:
Viewing 1 - 2 of 2 replies

Log in to reply.

Original Post
0 of 0 posts June 2018