Bucket A and block C are connected by a cable that passes...

Bucket $A$ and block $C$ are connected by a cable that passes over drum $B$. Knowing that drum $B$ rotates slowly counterclockwise and that the coefficients of friction at all surfaces are $\mu_{s}=0.35$ and $\mu_{k}=0.25,$ determine the smallest combined mass $m$ of the bucket and its contents for which block $C$ will $(a)$ remain at rest, $(b)$ start moving up the incline, $(c)$ continue moving up the incline at a constant speed.

Key Concepts

Force Decomposition on an Incline

This concept involves breaking a gravitational force into components that are parallel and perpendicular to an inclined surface. The parallel component causes motion along the incline, while the perpendicular component determines the normal force, which in turn is used to calculate frictional forces.

Static and Kinetic Friction

Static friction is the force that resists the initiation of motion between contacting surfaces until a maximum threshold is reached, as determined by the static friction coefficient. Once movement begins, kinetic friction, generally lower than static friction, governs the resistance to ongoing motion, both of which depend on the normal force.

Newton's Laws of Motion

Newton's laws underpin the analysis of both equilibrium and dynamic scenarios. They allow for the establishment of balance conditions when objects are at rest and relate net forces to acceleration when objects begin moving, forming the basis for analyzing motion and acceleration in connected systems.

Equilibrium Analysis

Equilibrium analysis involves setting the sum of forces (and torques, when applicable) equal to zero in order to determine the conditions for a system to remain at rest. This approach is essential to find the threshold condition at which an external force overcome static friction or other resistive forces.

Tension in Cable Systems

In systems where objects are connected by cables passing over pulleys or drums, the tension is a crucial variable that transmits force across different parts of the system. The analysis typically assumes either uniform tension or accounts for variations due to friction or rotational inertia in the connecting elements.

Transcript

00:02 As shown in the figure, vercat a and block c are connected by the string, passing over the drum b.

00:11 Mass of the block c is given 100 kg.

00:16 Cofficient of friction between the surfaces.

00:19 Static is 0 .35 and kinetic is 0 .25.

00:28 We have to find the smallest value of mass of a quackert and its content.

00:45 M we have to find a part for block c remains at rest b start moving up on the incline c continue moving up on the incline let us see a free body diagram of drum b since motion impending so t2 upon mg is equal to mu k into veter.

03:40 So t2 you will get exponential of 0 .25 to pi by 3 into mg that is 2 .0 814mg.

04:04 Now free body diagram of the block c, t2mg mg normal 30 degree this is friction mu s m g submission of force along this direction must be 0 so n minus mcg cause of 30 is equal to 0 so normal reaction you will get mcg cause of 30 so friction you will get mu s into n that is 0 .35 m g cos of 30 now summation of force along this direction must be 0 so you will get t 2 minus f minus mcg sign of 30 must be 0 value of t 2b have measured 2 .0814mg minus f is 0 .35 mcg cost 30 minus mcg sign of 30 must be 0 substituting the value 2 .0814 m into 9 .8 .8.

06:33 0 .35 100 into 9 .8 root 3 by 2 100 into 9 .8 so value of m you will get 9 .46 kg block c starts moving up meter of 120 degree so from here you will get t2 upon t1 exponential of mu as into meter so t2 is cut to mg exponential of m g exponential of point 48 hence t1 is called to point 480m g block c motion impends submission of force along this direction must be zero so normal reaction mcg cost of 30 so friction will be mu s into n that is point 35 mcg cost of 30 now taking submission of force along this direction that must be also zero so t1 minus f minus mcg sign of 30 must be zero substitute the value t1 is 048 045 mg f is 0 .35 mcg and cg is 035 mcg cause of 30 minus mcg half is equal to 0.

11:26 0 .48 .045m into 9 .8.

11:32 0 .35 mc that is 100...

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