Question

A uniform rod of mass 100g has a length of1m. It is supported horizontally on two knife edgesplaced 10cm from its end. What will be thereactions at these supports when a 50g mass issuspended 30cm from one of the knife edges?

Answer 1

We have to make the conversations

mass rod=100g=.1 kg

mass =50g=0.05 kg

in order to obtain the weight, we need to multiply the mass by the gravity in this case 9.8 m/s^2

w_rod= .1*9.8=0.98 N

w_mass=.05*9.8=0.049N

Then wee need to draw a diagram

We have RA and RB the reactions of the support

First we need to find the sum of forces

$\sum ^{\square}_{\square}F_y=RA+RB-P1-P2=0$

Then we analyze the moments of force

$\sum ^{\square}_{\square}M_{RA}=P1(0.3)+P2(0.4)-RB(0.8)=0$

then we can calculate RB

$RB=\frac{(0.49\cdot0.3)+(0.98\cdot0.4)}{0.8}=0.67375N$

then we can calculate RA

$RA=P1+P2-RB=0.49+0.98-0.67375=0.79625N$

the reactions will be

RA=0.79625N=0.80N

RB=0.67375N=0.67N