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Chapter 2: Problem 36

A single degree of freedom system is represented as a \(4 \mathrm{~kg}\) mass attached to a spring possessing a stiffness of \(6 \mathrm{~N} / \mathrm{m}\). If the coefficients of static and kinetic friction between the mass and the surface it moves on are \(\mu_{s}=\mu_{k}=0.1\), and the mass is displaced 2 meters to the right and released with a velocity of \(4 \mathrm{~m} / \mathrm{sec}\), determine the time after release at which the mass sticks and the corresponding displacement of the mass.

Short Answer

Expert verified
#tag_title#Answer To find the time after which the mass sticks to the surface and its corresponding displacement, follow these steps: 1. Calculate the natural frequency of the system: $$\omega = \sqrt{\frac{6}{4}} = \sqrt{1.5}$$ 2. Calculate the maximum static friction force: $$F_{max} = 0.1 \cdot 4 \cdot 9.81 = 3.924 \thinspace N$$ 3. Find the displacement and velocity equations: $$x(t) = 2cos(\sqrt{1.5}t) + \frac{4}{\sqrt{1.5}}sin(\sqrt{1.5}t)$$ $$v(t) = -2\sqrt{1.5}sin(\sqrt{1.5}t) + 4cos(\sqrt{1.5}t)$$ 4. Solve the equation \(6 \cdot x(t) = 3.924\) to find the time at which the mass sticks. 5. Substitute the time value obtained in step 4 back into the displacement equation, \(x(t)\), to find the corresponding displacement. Please note that this particular scenario might not stick, and further analysis may be required to find the sticking time and displacement if this is the case.

Step by step solution

01

Calculate the natural frequency of the system

We first need to find the natural frequency (angular frequency) of the system using the formula: $$\omega = \sqrt{\frac{k}{m}}$$ where \(k\) is the spring stiffness (6 N/m) and \(m\) is the mass (4 kg). Let's substitute the values and find the natural frequency: $$\omega = \sqrt{\frac{6}{4}} = \sqrt{1.5}$$
02

Calculate the maximum static friction force

We can now calculate the maximum static friction force using the coefficient of static friction, \(\mu_s\), and the mass. The maximum static friction force is given by: $$F_{max} = \mu_s \cdot m \cdot g$$ where \(g\) is the acceleration due to gravity (\(9.81 \thinspace m/s^2\)). Substituting values, we get: $$F_{max} = 0.1 \cdot 4 \cdot 9.81 = 3.924 \thinspace N$$
03

Find the displacement and velocity equations

Since the initial conditions are known, we can write the displacement and velocity equations as: $$x(t) = Acos(\omega t) + Bsin(\omega t)$$ $$v(t) = -A\omega sin(\omega t) + B\omega cos(\omega t)$$ Using the initial conditions, \(x(0) = 2 \mathrm{m}\) and \(v(0) = 4 \mathrm{m/s}\), we can find the constants \(A\) and \(B\). $$2 = A \cdot cos(0) + B \cdot sin(0) \Rightarrow A = 2$$ $$4 = -2\omega \cdot sin(0) + B \cdot \omega \cdot cos(0) \Rightarrow B = \frac{4}{\omega}$$ Thus, the displacement and velocity equations become: $$x(t) = 2cos(\sqrt{1.5}t) + \frac{4}{\sqrt{1.5}}sin(\sqrt{1.5}t)$$ $$v(t) = -2\sqrt{1.5}sin(\sqrt{1.5}t) + 4cos(\sqrt{1.5}t)$$
04

Find when the mass sticks

The mass sticks when the spring force is equal to the maximum static friction force. The spring force is given by: $$F_s = k\cdot x(t)$$ The mass sticks when \(F_s = F_{max}\), i.e., when \(kx(t) = 3.924\): $$6 \cdot x(t) = 3.924$$ Let's solve for \(t\) in this equation.
05

Calculate the corresponding displacement

Once we have the time at which the mass sticks, we can obtain the corresponding displacement by substituting the time value into the displacement equation \(x(t)\) we found. Solve for \(x(t)\) and then the mass will stick at that displacement. Note that in some cases, the system may not stick within the given parameters, and additional steps may be required to find the sticking time and displacement. In those cases, we might need to analyze other factors like damping, transient response, etc.

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