Two blocks of masses 3 kg and 2 kg are placed side by side on an incline as shown in the Fig.

Two blocks of masses 3 kg and 2 kg are placed side by side...

Two blocks of masses $3 \mathrm{~kg}$ and $2 \mathrm{~kg}$ are placed side by side on an incline as shown in the Fig. $7.418$. A force, $F=20 \mathrm{~N}$ is acting on $2 \mathrm{~kg}$ block along the incline. The coefficient of friction between the block and the incline is same and equal to $0.1$. find the normal contact force exerted by $2 \mathrm{~kg}$ block on $3 \mathrm{~kg}$ block. a. $18 \mathrm{~N}$ b. $30 \mathrm{~N}$ c. $12 \mathrm{~N}$ d. $27.6 \mathrm{~N}$

Two blocks of masses $3 \mathrm{~kg}$ and $2 \mathrm{~kg}$ are placed side by side on an incline as shown in the Fig. $7.418$. A force, $F=20 \mathrm{~N}$ is acting on $2 \mathrm{~kg}$ block along the incline. The coefficient of friction between the block and the incline is same and equal to $0.1$. find the normal contact force exerted by $2 \mathrm{~kg}$ block on $3 \mathrm{~kg}$ block.
a. $18 \mathrm{~N}$
b. $30 \mathrm{~N}$
c. $12 \mathrm{~N}$
d. $27.6 \mathrm{~N}$

Key Concepts

Newton's Second Law

This fundamental principle states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. It provides the basis for analyzing the dynamics of systems, including objects on inclined planes, where forces can be decomposed into components along different directions.

Inclined Plane Dynamics

When an object is on an inclined surface, the gravitational force can be decomposed into two components—one parallel to the incline that causes acceleration and one perpendicular that determines the normal force. Understanding these components is crucial for solving problems involving motion on slopes.

Friction

Friction is a resistive force that acts opposite to the direction of motion. It is typically modeled as proportional to the normal force acting between two surfaces through the coefficient of friction. Analyzing frictional forces is essential when calculating net forces and accelerations in systems involving contact between surfaces.

Free-Body Diagrams

These diagrams are used to visually represent all the forces acting on a single object. By isolating an object and depicting forces like gravity, normal force, applied forces, and friction, free-body diagrams help in setting up the equations needed to analyze motion and interactions between objects.

Contact Force Analysis

In systems where multiple objects interact, such as blocks in contact, it is important to analyze the forces they exert on each other. Contact force analysis involves applying Newton's laws to each object individually and utilizing the constraints of the system, such as common acceleration, to determine the magnitude and direction of the forces exchanged at the interface.

Transcript

00:01 Hello students in this question we have two blocks of mass m1 equals to 3 kg m2 equals to 2 kg are placed on the side by side on an inclined plane the force f equals to 20 newton is acting on the 2kg block okay so we have to calculate the normal contact force between the two blocks okay so let us suppose this is the inclined plane okay and we have the two blocks so suppose this is the first block and this is the second block and we are drawing here free body diagram so we have separated them okay so this is the normal force and this is the normal force okay and this will be here in the downward direction weight so this component of weight will be in this direction that is 3g and the angle with the inclination is 37 so cause sorry sign 37 degree okay sign 37 degree okay now the friction force f1 will act in this direction and that will be equals to mu and this one this is the n1 and that will be equals to 3g cost 37 okay and similarly the width component here will be 2g this is the weight of 2 kilogram so 2g sine 37 degree and the remaining values are here this is the friction force f2 and that will be equals to mu multiplied by 2g cost 37 and this is the applied force f here okay now both block will have acceleration in this direction so here we can calculate that the acceleration will be equals to net force divided by the net mass okay so here the net downward force will be this force plus this force and plus this force minus these two friction forces okay so 3g sign 37 degree plus 2g sign 37 degree plus f minus mu or 3 mu g cos 37 minus 2 mu g cos 37 and divide by the total mass that is equals to 3 plus 2 so acceleration after solving this will be equals to 46 divided by 5 meter per second square.

02:31 Now we have obtained the acceleration a, so we can apply the equation for any blog.

02:37 So we can apply for this blog.

02:40 So we can write here that 3g sign 37 degree plus n minus f1 that will be equals to mass m, that is here 3kg multiple at the acceleration a...

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