A 10N meterstick is suspended by two sping scales, one at the .08 m mark and the other at the .9 m mark. If a weight of 5N is hung at the .2 m mark and a weight of 17N is hung at the .55 m mark, what will be the reading on each scale?
I have:
F(_1)+F(_2)=5+10+17=32
With moment at F(_1), sum of torques=0:
F(_2)(.82)-5(.2)-10(.5)-17(.55)=0
.82F(_2)-1-5-9.35=0
.82F(_2)=15.35
F(_2)=18.72
This makes F(_1)=13.28
Do I now use F=ma as follows to find the scale reading?
F(_1)=m(_1)a
13.28=m(_1)(9.8)
m(_1)=1.355 or about 1.4 kg
12 3
To find the reading on each spring scale, you can use the equation F = kx, where F is the force exerted by the spring scale, k is the spring constant, and x is the displacement from the equilibrium position. The spring constant can be determined by dividing the force applied to the spring scale by the displacement.
First, let's calculate the spring constant for the spring scale at the 0.08 m mark. The displacement is 0.08 m and the force is 13.28 N:
k₁ = F₁ / x₁
k₁ = 13.28 N / 0.08 m
k₁ ≈ 166 N/m
Now, let's calculate the reading on the spring scale at the 0.08 m mark:
F₁ = k₁ * x₁
F₁ = 166 N/m * 0.08 m
F₁ ≈ 13.28 N
So, the reading on the spring scale at the 0.08 m mark is approximately 13.28 N.
Next, let's calculate the spring constant for the spring scale at the 0.9 m mark. The displacement is 0.9 m and the force is 18.72 N:
k₂ = F₂ / x₂
k₂ = 18.72 N / 0.9 m
k₂ ≈ 20.8 N/m
Now, let's calculate the reading on the spring scale at the 0.9 m mark:
F₂ = k₂ * x₂
F₂ = 20.8 N/m * 0.9 m
F₂ ≈ 18.72 N
So, the reading on the spring scale at the 0.9 m mark is approximately 18.72 N.
To summarize, the reading on the spring scale at the 0.08 m mark is approximately 13.28 N and the reading on the spring scale at the 0.9 m mark is approximately 18.72 N.