Figure 6-14 shows a block of mass m connected to a block of mass M 2.00 kg, both on 45 inclined

Figure 6-14 shows a block of mass m connected to a block of...

Key Concepts

Newton's Laws in Equilibrium

This concept involves applying Newton's laws to systems at rest, where the net force on each component is zero. In static equilibrium problems, forces such as gravity, friction, normal, and tension must balance so that there is no acceleration.

Static Friction

Static friction prevents motion between surfaces in contact until a threshold force is applied. Its magnitude is determined by the coefficient of static friction multiplied by the normal force, and it must be overcome for motion to occur.

Inclined Plane Dynamics

When analyzing forces on an inclined plane, the weight of an object is decomposed into components parallel and perpendicular to the plane. The parallel component affects the motion along the plane, while the perpendicular component influences the normal force and therefore the friction.

Free Body Diagram Analysis

Drawing free body diagrams is crucial for visualizing and resolving all the forces acting on each mass. This technique helps in setting up the equilibrium equations by clearly identifying gravitational, normal, frictional, and tension forces acting on the system.

Transcript

00:01 In the question it is given that the mass of block m is equal to 2 kg and the coefficient of static friction between the surface of the plane and the block are 0 .28.

00:16 Also the angle of inclination of the plane is given as 45 degrees.

00:22 So for the first part of the question we are asked to find out the minimum value of the plane.

00:31 Small m for which the system will be in static condition so let us resolve the free body diagram of the respective blocks so this is the direction of the normal reaction and this is the direction of the weight which is m g for this block so along this direction will be the sign component which is mg sine theta and and this direction will be the cost component which is m g cost theta and this is the direction of the tension on the rope similarly for the second block this is the direction of the tension on the rope let this be the normal reaction and let us mark it as n dash so this is the direction of the weight which is small m multiplied by g this is the cost component of m g which is m g which is m g which is m g cost theta and this is the sign component of the small mass m which is m g sine theta so for the smaller mass m we can write t is equal to the force of friction fs plus mg sine theta so this is the direction of force of friction f s for for this block and for this block this is the direction of force of friction which is f s now we know f dash s can be written as mu multiplied by n1 dash or the normal reaction of the block plus mg sine 45 degree which is equals to mu m g cost 45 degree plus mg sine 45 degree taking m g and we know cost 45 is equal to sine 45 is equal to 1 by root 2 we can take the entire thing as common we get m g by root 2 in common 1 plus mu which is equal to the value of d let us mark this equation as number 1 now for this second block of bigger which is m we can write t plus the friction force which is f s is equals to the cost component of the weight of the block which is m g cost 45 degree or we can write t plus f s which is equals to mu multiplied by n is equals to 2g cost 45 degree of putting the value of n we get t is equal to 2g cos 45 degree minus mu 2g sine 45 degree which is equal to taking 2g and cos 40 by sign 45 degree as common we get 2g by root 2 1 minus mu is equal to taking 2g and cos 40 by 5 degree is to the value of t so let us mark this equation as equation number two now equating equation number one to equation number two we get m g by root 2 1 plus mu is equals to 2g by root 2 1 minus mu so cancelling g and root 2 from both side or we get m is equal to 2 multiplied by 1 minus new divided by 1 plus mu now putting the value of mu we get to 1 minus 0 .28 divided by 1 plus 0 .28 which is equals to 1 .125 newton so this is the minimum value of m for which this is will be in static condition.

05:32 Now for the second part of the question we are asked to find out the maximum value of m for which the system will be in rest...

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