Two springs and two masses are attached in a straight vertical line as shown in Figure Q3. The system is set in motion by holding the mass m₂ at its equilibrium position and pushing the mass m₁ downwards of its equilibrium position a distance 2 m and then releasing both masses. if m₁ = m₂ = 1 kg, k₁ = 3 N/m and k₂ = 2 N/m. www.m k₁ = 3 (y₁ = 0). m₁ = 1 k2=2 (y₂ = 0) |m₂ = 1 Y2 y 2 System in static equilibrium (Net change in spring length =32-31) System in motion Figure Q3 - Coupled mass-spring system Determine the equations of motion y₁(t) and y₂(t) for the two masses m₁ and m₂ respectively: Analytically (hand calculations)

Two springs and two masses are attached in a straight vertical line as shown in Figure Q3. The system is set in motion by holding the mass m₂ at its equilibrium position and pushing the mass m₁ downwards of its equilibrium position a distance 2 m and then releasing both masses. if m₁ = m₂ = 1 kg, k₁ = 3 N/m and k₂ = 2 N/m. www.m k₁ = 3 (y₁ = 0). m₁ = 1 k2=2 (y₂ = 0) |m₂ = 1 Y2 y 2 System in static equilibrium (Net change in spring length =32-31) System in motion Figure Q3 - Coupled mass-spring system Determine the equations of motion y₁(t) and y₂(t) for the two masses m₁ and m₂ respectively: Analytically (hand calculations)

International Edition---engineering Mechanics: Statics, 4th Edition

4th Edition

ISBN:9781305501607

Author:Andrew Pytel And Jaan Kiusalaas

Publisher:Andrew Pytel And Jaan Kiusalaas

Chapter10: Virtual Work And Potential Energy

Section: Chapter Questions

Problem 10.57P: Find the stable equilibrium position of the system described in Prob. 10.56 if m = 2.06 kg.

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Two springs and two masses are attached in a straight vertical line as shown in Figure Q3. The system is set in motion by holding the mass m₂ at its equilibrium position and pushing the mass m₁ downwards of its equilibrium position a distance 2 m and then releasing both masses. if m₁ = m₂ = 1 kg, k₁ = 3 N/m and k₂ = 2 N/m. www.m k₁ = 3 (y₁ = 0). m₁ = 1 k2=2 (y₂ = 0) |m₂ = 1 Y2 y 2 System in static equilibrium (Net change in spring length =32-31) System in motion Figure Q3 - Coupled mass-spring system Determine the equations of motion y₁(t) and y₂(t) for the two masses m₁ and m₂ respectively: Analytically (hand calculations)
Two springs and two masses are attached in a straight vertical line as shown in Figure Q3. The system is set in motion by holding the mass m₂ at its equilibrium position and pushing the mass m₁ downwards of its equilibrium position a distance 2 m and then releasing both masses. if m₁ = m² = 1 kg, k₁ = 3 N/m and k₂ = 2 N/m. (y₁ = 0) www k₁ = 3 Jm₁ = 1 k2=2 www (Net change in spring length =32-31) (y₂ = 0) m₂ = 1 32 32 System in static equilibrium System in motion Figure Q3 - Coupled mass-spring system Determine the equations of motion y₁ (t) and y₂(t) for the two masses m₁ and m₂ respectively: Analytically (hand calculations) Using MATLAB Numerical Functions (ode45) Creating Simulink Model Produce an animation of the system for all solutions for the first minute.
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1. The PASCO human arm model is configured such that the cord representing the bicep is perfectly vertical and the forearm is at 90° (in the figure to the right, the cord is not quite vertical). A mass of 100 g is attached to the hand. Draw a free-body diagram on the figure to the right showing all forces which act on the forearm. The force of the bicep F on the arm The force of the humerus FH on the arm The weight of the forearm W The mass in the hand Wm 100 g Be careful to draw the force vectors with tails beginning at the point where the force is actually applied to the forearm. 2. Consider the free body diagram below. Determine the perpendicular component F̟ of the force F exerted by the biceps brachii on the forearm. Use the fact that cos 0 = H/B to write this component directly in terms of the humerus length H and the biceps length B. H 3. If the forearm is in equilibrium, then there is no angular acceleration and therefore the sum of the torques applied to the forearm must be…
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