How Do Weight Distribution and Torque Affect Scale Readings?

AI Thread Summary
The discussion revolves around calculating the scale readings when a woman stands on a light board supported by two scales. The woman's weight is converted to 693.3 N, and the distances from the scales are used to determine the readings. The calculations initially yield 559.97 N for one scale and 133.33 N for the other, but the results are incorrect according to the computer. Participants suggest reevaluating the moments about one scale to find the correct distances and reactions. The importance of considering the board's mass and accurately applying torque principles is emphasized for accurate scale readings.
jibjab
Messages
13
Reaction score
0

Homework Statement


A 70.7 kg woman stands on a very light, rigid board that rests on a bathroom scale at each end. The total length of the board is L=2.1 m, and the woman's distance from the left of the board is 1.3 m.
What is the reading on the left scale? The right scale?

Homework Equations



n2=d1w/d2


The Attempt at a Solution


I converted the woman's weight to 693.3 N. For d1 I used half the length of the board, 1.05 m, and for d2, I found the distance the woman was away from the right side of the board relevant to the center of gravity on the board, which was 0.25 m. To find n2, I did
(693.3N)(1.05m)/1.3m and got 559.97 N.
I subtracted that amount from the woman's weight and got 133.33 N, which made sense since she was closer to the right side of the board, but the computer says no. Can anyone see what I'm doing wrong? I was following an example in the book.

Thank you!
 
Physics news on Phys.org
She is standing on a very light board. No neglect its mass.
Let R1 and R2 are the reactions on the scales. Find the moments due to woman and R2 about R1. Hence find the distance of the woman from R1.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top