Calculate the vertical acceleration of the system mass center and the individual vertical accele

Calculate the vertical acceleration of the system mass...

Key Concepts

Center of Mass Acceleration

The center of mass of a system is the weighted average position of all its masses. Its acceleration is determined solely by the net external force acting on the system divided by the total mass (as per Newton’s second law). This concept is crucial in dynamics for reducing a complex system to a single point whose motion represents the aggregate effect of all forces.

Newton's Second Law

Newton’s second law establishes the fundamental relationship between force, mass, and acceleration (F = ma). This principle allows one to calculate the acceleration of bodies when forces are applied, which is essential for analyzing both the overall system response and the individual components in dynamic problems.

Free-Body Diagrams

Free-body diagrams are graphical representations used to illustrate all forces acting on a particle or component. They are important for systematically breaking down the forces, especially in systems with multiple parts connected by wires or rods, and for applying Newton’s laws accurately to each part of the system.

Symmetry and Force Distribution

In problems where symmetrical conditions (such as equal angles or geometry) are present, such symmetry simplifies the analysis by leading to an equal distribution of forces across components. When symmetry is absent, as in the case when angles differ, the forces and resulting accelerations must be determined separately for each component, reflecting non-uniform loading conditions.

Assumptions in Idealized Mechanical Systems

Many dynamic analyses assume ideal conditions such as massless connectors, frictionless pivots, and neglect of air resistance. These assumptions simplify the system by focusing on the main forces and responses, facilitating the application of theoretical principles without the complications of real-world imperfections.

Transcript

00:01 So here in this problem we are given with these two sphere of mass 2 kg so mass of sphere 1 is equals to mass of sphere 2 and that is equals to 2kg and they are attached with these two wire and at this point a force that has a magnitude of 15 newton is being applied now with there basically two cases we have to assume that in the first case alpha is equals to beta this angle are equal and then we have to calculate the vertical axel for the sphere 1 and 2 and for the system now next in the case b we have to consider that alpha is not equals to beta and then we have to calculate a vertical acceleration of the system and the two sphere separately to solve this problem we'll be considering the first case where alpha is equals to beta so this is case 1 alpha is equals to beta and in this case we'll be first calculating the vertical acceleration for the system for that we can use a nutrient second law of motion for the system of bodies and which is sigma f net is equals to mass times the acceleration this is a low now the net force if you consider this as our whole system so net force acting on the system is we are considering the force in the vertically upward direction to be positive and the vertically downward direction to be negative so here is m1g would be acting in the downward direction and here m2g would be acting in the downward direction now we can apply this low so f1 minus m1g minus m2g is equals to m1 plus m2 times the vertical acceleration a y we've applied this low in the vertical direction so we've obtained this equation now from here we can easily calculate so this f1 is basically this if if only so we can write if only if we have 50 minus m1 is two times g is acceleration to gravity its value is 9 .81 minus 2 times 9 .81 is equals to m1 plus m2 would become 4 times a y so from here we'll get the value of a y to be nearly 2 .6 9 meter per second squared so this is the acceleration for the system now next we have to calculate the acceleration of the individual spheres now the acceleration of the individual sphere a y1 that is the acceleration of the sphere one it is equals to acceleration of the sphere 2 and that will be equals to 2 .69 meter per second squared the acceleration would be same because of the symbol of the system so due to symmetricity the acceleration would be same now next i will be considering the v case where alpha is not equals to beta so in this part we have to consider that alpha is not equals to beta and then we have to calculate the acceleration for the system in the vertical direction and for the individual sphere also so first it will be calculating the acceleration of the center of mass by considering as a system at this as a whole as a system and we can apply newton's second law of motion here which is sigma f net we are applying in the vertical direction is equals to mass times the acceleration so if net we have f minus m1 plus m2 times g is equals to m2 plus m1 times a y now we have all the values you can plug it here mass the force we have 15 newton m1 and m2 is 4 times g is 9 .81 is equal to 4 times a y so from here a y comes out to be as 2 .69 meter per second squared so this is the acceleration for the system when alpha is not equals to beta next we'll be considering the will be finding the acceleration for individual spheres so in the b case, alpha is not equals to beta.

05:05 Now let the tension force in this cable be t1 and in this cable let it be t2.

05:14 Now we'll be calculating the expression for t1 and t2 in terms of the force if the given data, that is the force if in the angle alpha and beta...

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