Statics - long rod of linear varying density supported at both ends

In summary, the conversation discusses a 3.0 m long rod with a density described by a + bx, where a = 5.0 kg/m, b = 5.0 kg/m2, and x is the distance in meters from the left end of the rod. The rod is supported by two scales and the conversation explores how to find the readings of these scales using the mass and location of the center of mass of the rod.
  • #1
jtulibarri
2
0
A 3.0 m long rod has a density described by = a + bx, where is the density in kilograms per meter of length, a = 5.0 kg/m, b = 5.0 kg/m2, and x is the distance in meters from the left end of the rod. The rod rests horizontally with its ends each supported by a scale. What do the two scales read?


I don't even know how to start this one.

If anyone can show me how to set this up I can solve the rest
 
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  • #2
First find the mass of the rod, and the location of the center of mass of the rod...

The measurement the scale reads, is the magnitude of the normal force that the scale exerts...

You can find one of these forces using net torque about the appropriate point = 0...

Then the other force can also be obtained using torque... or using sum of forces in y-direction = 0.
 
  • #3
.

Sure, I can help you set this up. First, let's define the variables and their units:

L = 3.0 m (length of the rod)
x = distance in meters from the left end of the rod
ρ = density in kg/m of length
a = 5.0 kg/m (constant)
b = 5.0 kg/m2 (constant)

Now, we can use the definition of density to find the mass of a small section of the rod at a distance x from the left end:

m = ρ * dx

Where dx represents a small length element of the rod. We can then integrate this equation over the length of the rod to find the total mass:

m_total = ∫ρ * dx from x = 0 to x = L

Since the density varies along the length of the rod, we need to express ρ in terms of x:

ρ = a + bx

Substituting this into the integral, we get:

m_total = ∫(a + bx) * dx from x = 0 to x = L

Integrating, we get:

m_total = ax + (b/2)x^2 from x = 0 to x = L

m_total = aL + (b/2)L^2

Now, we can use this equation to find the total mass of the rod, which will be equal to the sum of the masses supported by the two scales:

m_total = m_scale1 + m_scale2

Since the rod is resting horizontally, we can assume that the weight of the rod is evenly distributed between the two scales. Therefore, we can set up the following equations:

m_scale1 = (m_total/2) = (aL + (b/2)L^2)/2

m_scale2 = (m_total/2) = (aL + (b/2)L^2)/2

Now, we can plug in the values for a, b, and L and solve for the masses supported by each scale:

m_scale1 = (5.0 kg/m * 3.0 m + (5.0 kg/m2/2) * (3.0 m)^2)/2 = 22.5 kg

m_scale2 = (5.0 kg/m * 3.0 m + (5.0 kg/m2/2) * (3.0 m)^2)/
 

1. What is the definition of "statics"?

Statics is the branch of mechanics that studies the behavior of objects at rest, or in equilibrium, under the action of external forces.

2. How is the density of a long rod with linear varying density typically described?

The density of a long rod with linear varying density is typically described using a function, where the density varies with respect to the length of the rod.

3. How are the forces acting on a long rod with linear varying density determined?

The forces acting on a long rod with linear varying density are determined by the weight of the rod and the external forces acting on it, such as supports or applied loads.

4. What are the boundary conditions for a long rod with linear varying density supported at both ends?

The boundary conditions for a long rod with linear varying density supported at both ends are that the rod is fixed and cannot move at both ends, and that the sum of the forces acting on the rod must be equal to zero at each support.

5. What is the significance of studying the statics of a long rod with linear varying density?

Studying the statics of a long rod with linear varying density allows us to understand the behavior of objects under various external forces and to design structures that can support these forces without failing. It also helps us to predict and prevent potential failures or deformations in real-life structures.

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