Block B in Fig. 6-31 weighs 750 N. The coefficient of static friction between block and table is

Block B in Fig. 6-31 weighs 750 N. The coefficient of static...

Key Concepts

Static Friction

Static friction is the resisting force that prevents relative motion between two surfaces in contact. It must be overcome by any applied force for motion to begin, and its maximum value is typically calculated using the product of the coefficient of static friction and the normal force between the surfaces.

Force Equilibrium

Force equilibrium refers to the condition in which the net force acting on an object is zero, meaning that all forces are balanced. This concept is crucial in statics problems where the sum of forces acting in any direction must be equal to zero to ensure the system is stationary.

Tension in Ropes

Tension in a rope or cable is the pulling force transmitted along its length when forces are applied at its ends. In problems involving interconnected masses by ropes, tension is a key variable that transmits the forces between the different parts of the system.

Free Body Diagram

A free body diagram is a simplified graphical representation that depicts all the forces acting on a single object. It is an essential tool in mechanics that helps in the systematic analysis of forces, such as weight, friction, tension, and applied forces, to solve equilibrium problems.

Force Resolution

Force resolution involves breaking down a force into its components along specific axes, usually horizontal and vertical. This process is fundamental in analyzing influences of forces applied at angles, allowing for a clearer understanding of the net effect in each direction.

Transcript

00:01 In this question it is given that the weight of the block b is given as wb is equal to 750 newton and the coefficient of static friction between the block b and the debit is mu s is equal to 0 .25 and the angle theta is equal to 30 degree so we are asked to find out the maximum weight of the block a for which the system is will be stationary so for stationary condition we know the acceleration of the system will be equals to 0 meters per second square so let us resolve the different forces acting on the body so this is the direction of the normal reaction and this is the direction of the weight wb of the block b this is the direction of the weight w b this is the direction of the weight wa of the block a so let this be tension t1 and this be tension t2 so along this direction we get t2 cost theta as the value of tension or from here we can see t1 is equal to t2 cost theta also the friction force acting on the block b is given as f s and we know f s is equal to mu times the normal reaction is equals to mu multiplied by w b so for block b we can write the equation t1 minus fs is equal to mass of the block b multiplied by its acceleration this is according to the newton's second block since the acceleration is zero as the system is in static condition we can write a is equal to zero so we get the integration equals to zero and for the second block which is block a we can write the equation so along this direction is equal to t to sine theta so this is the tension or we can write t2 sin theta minus w a is equal to m a multiplied by a which is equal to 0 now from the equation of block b it can be resolved as t1 is equal to f s max or the maximum friction and from these equation of block a we can write t 2 sine theta is equal to w a or weight of the block b so now let us mark this equation as equation number 1 this one as equation number 2 this one as equation number 3 and this one as equation number 4 so from equation number 1 we get t 1 is equal to t 2 cost or from here we get t2 is equal to t1 by cost theta so now let us take it equation number 5 now putting equation number 5 and equation number 3 in equation number 4 we get fs max by cos theta multiplied by sine theta is equal to w a or weight of the block b or now putting the values of the respective quantity we can write 0 .25 multiplied by n which is greater of the block b which is 750 newton multiplied by 10 30 degree is equal to w a or from here we get the value of w .a is equal to 108 .25 newton...

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