Two masses 40 kg and 30 kg are connected by a massless string passing over a frictionless ligh

Two masses 40 kg and 30 kg are connected by a massless...

Key Concepts

Newton's Second Law

This fundamental principle of mechanics states that the net force acting on an object is equal to the product of its mass and its acceleration (F = ma). It provides the basis for setting up the equations of motion for systems of masses, allowing one to solve for unknown quantities such as acceleration by balancing all the forces acting on the system.

System Approach

When dealing with interconnected objects, it is often useful to consider the entire system as a single body. In this approach, internal forces (like the tension in a string) cancel out, and the net external force acting on the combined mass is used to determine the acceleration. This simplifies the analysis and avoids having to consider the interaction forces between individual objects.

Free-Body Diagram

A free-body diagram is a schematic representation used to visualize all the forces acting on each component of a system. By isolating an object and drawing vectors for gravitational forces, tension, friction (if any), and other applied forces, one can systematically apply Newton's laws to each mass. This method is crucial for correctly setting up the equations that govern the motion of the system.

Atwood Machine Concept

An Atwood machine is a classic physics problem that involves two masses connected by a string over a pulley. It is a fundamental example used to illustrate the application of Newton's second law, the role of tension in a string, and how differences in mass lead to acceleration. The simplicity of the Atwood machine makes it an ideal model for understanding basic principles of dynamics.

Massless String and Frictionless Pulley Assumptions

These idealizations are commonly used in physics problems to simplify the analysis. A massless string ensures that the tension is the same throughout the string, while a frictionless and light pulley eliminates any rotational inertia or frictional forces. This means that only the gravitational forces and the masses' accelerations need to be considered, greatly simplifying the mathematical treatment of the problem.

Transcript

00:01 In this question, they have given that these two blocks are attached with the weightless string and they have given that these two blocks are inclined at an angle of 30 degree.

00:10 So we have to calculate the acceleration of this system.

00:13 So if i make a free body diagram for mass m2, so it will be like due to gravity there will be a downward force that will be m2g.

00:23 This is the component of this force that will be equal to m2g cause of theta and this will be the sign component of this force that will be m2g sign of theta and the tension is present in the string that will be t and let's say we are assuming that the acceleration it is in this direction right and if i make a free bubble diagram for m1 now so due to gravity there will be m1g and the component of this force will be equal to m1g cause of theta where theta is 30 degree the sign component will be in this direction that will be equal to m1g sine of theta and the tension will be in this direction and for this block m1 the acceleration will be in upward direction that is a now if i write the equations now for block m2 the equation will be m2g sine of theta minus tension t that will be mass multiply by acceleration because we know that the newton's law is equal to f net it will be equal to m multiply by acceleration so if i put the values now so my m2 is 30 kilogram multiplied by g sign of 30 degree minus tension it will be equal to mass that is mass 2 that will be 30 multiply by acceleration that we have to calculate now if i write the equation for mass m1 so i will write that the t minus m1 g sign of theta that will be equal to mass of 1 that will be multiplied by the acceleration this is our equation 2 and this is our equation 1 so if i add these two equation so my final equation will be 30 g sign of 30 degree minus m1 is given that is 40 multiply by g that is 30 degree it will be equal to 30 multiplied by a plus 40 multiply by a right now if i take g sign 30 degree common so it will be 9 .8 the value of g is 9 .8 multiply by sine 30 is 1 divided by 2 bracket 30 minus 40 it will be equal to 70a now if we will solve this we will get acceleration will be equal to minus 0 .7 meter per second square.

03:09 So this is the acceleration and this negative sign shows that we assume the acceleration in wrong direction...

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