A 10m long uniform beam is pivoted in its centre. A 30kg point mass is placed on one end of the beam. Where must a 50kg mass be placed in order to balance the beam?

For this question, we will use moments. A moment is defined as:Moment = force x perpendicular distance from pivotHere, the moment of the 30kg mass which acts anti-clockwise through the pivot. The moment is:M30kg = 30g x 5mThe force is 30g as F=ma with g being the acceleration under gravity. The perpendicular distance is 5m as the beam is pivoted in the centre and the mass is placed at the end of the beam.The moment of the 50kg mass is:M50kg = 50g x DWhere D is the distance from the pivot. Since we know for the beam to be balanced, the clockwise moment must be equal to the anti-clockwise moment, we can say:M30kg = M50kg30g x 5 = 50g x DWe can cancel out the g factor as it is present on both sides of the equation.30 x 5 = 50 x DD = (30 x 5)/50 = (30 x 1)/5 = 3mSo the mass must be placed 3m from the pivot which is also 8m from the end which the 30kg mass is placed on.

JM
Answered by Joseph M. Physics tutor

2731 Views

See similar Physics A Level tutors

Related Physics A Level answers

All answers ▸

A box initially at rest is on a plank, of length 5m, that is elevated at an angle such that tan(a)=3/4. When it reaches the end of the plank it has velocity 5ms^-1. Calculate the average frictional force on the box.


How do stars form?


A cannon can fire a cannonball at 20m/s. A sandpit is placed at a distance of 40m away. At what angle should the cannon be fired in order for the cannonball to land in the sand.


In the Photoelectric Effect, what is mean by 'threshold frequency' and how does the magnitude a photons frequency effect the electron it is absorbed by?


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences