equilibrium

[Addallat]

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I originally started this post having a tough time with this question, but wound up with the right answer by the time I was through working everything out (see my work done below).
Maybe I'm just not comfortable with torque Q's, but I thought this was a tough question. Do you guys feel this question is easy? It took me 3 minutes to solve which is too much time to spend on a discrete. I guess I just need more practice....

A Uniform plank of mass 12 kg and length L is positioned horizontally, with two ends supported by sensitive scales, an object of mass 3 kg is placed a distance L/3 from the left end of the plank. What weight does the right hand scale read?


A. 50 N
B. 70 N
C. 80 N
D. 90 N


1. I noticed that the object is static so the sum of the forces(torques) must be zero.


So from what I can recognize I see 3 torques

1. counter clockwise torque cause by the 3 kg mass that's 1/3 the distance from the left end.

so
Torque1: 30N(1/3)L


2. The counter clockwise torque caused by the plank itself:

Torque2: 120(1/2)L


3. The clockwise torque at the right side of the plank (The force of which will give you, your answer)
L(F2)


So what you wind up with is:

10 L + 60L - L (F) = 0

Solve for F = 70 N

Answer is B

 

[pm1]

It took me a while too..

My question is, there are two scales. Would they read the same thing? How would that work with having one scale supporting each end?

 

[FeinMS]

Using the left end as the pivot point
Torque clockwise:
1. by 3kg mass distance L/3 away from the left scale 3g(L/3)
2. by the center of mass of the plank distance L/2 away from the left scale 12g(L/2)
Torque counterclockwise:
1. by right hand scale distance L away: x(L)
Note: x is the reading on the scale.
3g(L/3)+12g(L/2)=x(L) solving for x gives 7g=70N