MemberJune 21, 2021 at 9:44 am968 Replies159150 XPContributions: EnterMedSchool Donor Team Manager VolunteerReputation: 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?
MemberJuly 15, 2021 at 10:51 pm984 Replies59071 XPContributions: VolunteerReputation: Level 6
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
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