A 400 -pound weight is suspended by two cables (see figure)....

A 400 -pound weight is suspended by two cables (see figure). What is the force (tension) on each cable? [Hint: Imagine that the weight is at the origin and that the dashed line is the $x$ -axis. Then cable $\mathbf{v}$ is represented by the vector $\left(c \cos 65^{\circ}\right) \mathbf{i}+\left(c \sin 65^{\circ}\right) \mathbf{j},$ which has magnitude $c$ (why?). Represent cable u similarly, denoting its magnitude by $d$ Use the fact that $\mathbf{u}+\mathbf{v}=0 \mathbf{i}+400 \mathbf{j}$ (why?) to set up a system of two equations in the unknowns $c \text { and } d .]$ CAN'T COPY THE FIGURE

Key Concepts

Static Equilibrium

This concept refers to a state where a system is at rest or moving at a constant velocity, meaning that the sum of all external forces and moments acting on the system is zero. In problems involving suspended weights, this principle ensures that the forces in different directions balance perfectly, leading to equations that help determine unknown forces.

Vector Resolution of Forces

This is the process of breaking down a force into its constituent components along chosen coordinate axes, typically horizontal and vertical directions. It is crucial in analyzing forces acting at angles, as it allows one to handle each component separately in the equilibrium equations.

Trigonometric Decomposition

Trigonometric functions, particularly sine and cosine, are used to express the horizontal and vertical components of a force vector when its magnitude and direction are known. This method is fundamental in converting the angled forces in the cables into usable component forms for further analysis.

System of Equations

When a physical situation involves multiple force components balancing out, the equilibrium conditions can be written as a system of linear equations. Solving these equations simultaneously provides the values of unknown quantities such as tensions or force magnitudes.

Transcript

00:01 We have problem number 73 in which a 400 pound weight is suspended by two cables so these are the two cables.

00:13 Okay, so let us we have to find the value of you and v that is tensions in each of the tension in each of the each of the cables due to this weight okay, so let us suppose you is the tension in cable number one and v is the tension in cable number two and let us set up coordinate system that is suppose this to be x and this to be y okay now u or v first of all v vector v can be represented can be resolved in two uh can be resolved in our two components that is vertical and horizontal so vertical and horizontal so horizontal component will be v cost 65 degree i cap because this will be in x direction plus v sign 65 degree and j similarly tension in cable u that is cable number one will be represented by can be resolved in rectangular components as u cost 32 degree but it will be in a post direction of x -axis so this will be minus i plus u sine 32 degree j now we have u plus v equal to 400 pounds okay 400 pounds so 400 can also be represented as 400 pounds so weight will be only in this direction okay so this will be 0 i 0 i plus 400 j okay 400 j with negative sign we should say okay okay now one more thing we should observe that here we need not to write 400 j as we are equating them equating them means we are already taking care of this so u plus v let us just plug in all these values 65 degree iota plus v sign 65 degree j plus we should write minus because this is minus iata minus u cost 32 degree iota plus u sine 32 degree j now let us take i common from these two so we will be having v cos 65 degree minus u cost 32 degree iata plus v sign 65 degree plus u sign 32 degree j equal to zero aita plus 400 j by equating the coefficients of i and j from both the sides this will be equal to zero that v cos 65 degree minus u cost 32 degree will be equal to zero and v sign we are equating this remember 65 degree plus u sign 32 degree will be equal to 400 okay so from here we can write as we cos 65 degree equal to u cost 32 degree so v will be equal to u cost 32 degree by cost 65 degrees okay now let us plug in here the value of b so v is u cost 32 degrees by cost 65 degrees into sign 65 degrees plus u sine 32 degrees equal to 400 now let us take u as common we will be getting sine 65 by cost 65 that is 1065 that it cost 32 degree 1065 degree plus sine 32 degrees equal to 400 so u will be equal to 400 divided by all these terms cost 32 degree 10 65 degrees plus sign 32 degrees now let us use calculator to get this so 400 divided by cost 32 degree cost 32 degree into k into 1065 degrees 1065 degrees plus sign 32 degrees okay so this will be equal to 170 .38 pounds.

06:21 Okay, so this is the value of you.

06:25 Now let us plug in here to get the value of v.

06:29 So, v will be equal to 170...

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