An elevator and its load have a combined mass of 1600 kg. Find the tension in the supporting ca

An elevator and its load have a combined mass of 1600 kg....

An elevator and its load have a combined mass of $1600 \mathrm{~kg}$. Find the tension in the supporting cable when the elevator, originally moving downward at $12.0 \mathrm{~m} / \mathrm{s}$, is brought to rest with constant acceleration in a distance of $42.0 \mathrm{~m}$.

Key Concepts

Kinematics of Uniform Acceleration

This concept involves using equations that describe motion with constant acceleration to relate variables such as initial velocity, final velocity, acceleration, and displacement. In problems like this, one often employs the relation v² = v?² + 2a?x to determine the acceleration when other quantities (initial velocity, final velocity, and displacement) are known.

Newton's Second Law of Motion

Newton’s Second Law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration (F_net = ma). This principle is fundamental when analyzing the forces in a system, as it links the calculated acceleration from kinematics to the forces, such as tension in a cable and gravitational force, acting on the mass.

Force Analysis in Accelerated Systems

This concept involves constructing a free-body diagram to identify all forces acting on an object, determining their directions, and writing a net force equation. When an object is accelerating, the sum of the forces (for example, tension upward minus the weight downward) must equate to the mass times the acceleration, which is crucial for correctly solving for unknown forces like the tension in a supporting cable.

Transcript

00:02 Hello student we are going to understand this question here given given combined mass of elevator and load let it is represented by m which is given 1600 kg tension in supporting cable that it is represented by t here to further tension in the supported cable so here given initial velocity u it is represented by by you which is given 12 meter per second and constant and distance travel by the elevator in the downward direction that is represented by d distance is represented by d which is 42 .0 meter so we have to find here we have to find the acceleration let the acceleration be represented by a that acceleration or de -accelation of elevator is represented by a is represented by is represented by a so we can write v square is equal to u square minus 2 a into s or 2 a into d here as is distance or 2a into d.

01:57 Here given final velocity of the elevator it is 0.

02:01 Initial velocity is given 12 minus 2 into value of a, we have to calculate and value of d it is given for 3 to meter.

02:12 Now solving a is equal to 1 1st2 x2.

02:21 Upon 42 into 2 so after solving we will get 12 .7 meter per second square now we will calculate the tension in the supported cable so let draw the block diagram tension in the cable t is equal to m g by minus m a because net acceleration is in downward direction with acceleration k so we can write t is equal to m into g minus m here we will take common m will be common so g minus k now substituting the value in this t is equal to value of m is sixteen hundred value of g 9 .8 meter per second square minus value of a that is 12 of 7 which we can write 1 .71 newton.

03:40 Now further solve it t is equal to 1600 and 9 .8 minus 1 .7 that is 8 .09 newton...

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