P = 1650 N, ? = 39°, ? = 55°, ? = 22° A load P is supported...

P = 1650 N, ? = 39°, ? = 55°, ? = 22° A load P is supported by two cables as shown. Considering the equilibrium at point C, estimate the tension in the cables. (Answer up to 1 decimal place) Tension in cable AC: ____ N. Tension in cable BC: ____ N. In vector form: The tension in cable AC: [____ i, ____ j] (N). The tension in cable BC: [____ i, ____ j] (N).

Transcript

00:01 So here we have these three cables or two cables and this extra force p.

00:08 And the cable a has an angle of 39 degrees with the horizontal.

00:13 Cable b has an angle of 55 degrees.

00:15 And p is being applied with an angle of 22 degrees with the vertical.

00:19 And we know that p has a value of 1650 or 1 ,650 newtons.

00:26 So in order to analyze this, the first thing we need to do is divide this into its x and y components.

00:33 And so looking at the x direction first, we are going to have the tension in the cable ac, which i'm going to label ta.

00:44 The horizontal component of that, ta cosine alpha, will equal the tension in cables b and the pressure or the force p, also in the x direction.

00:58 So tb, cosine beta plus p sine theta.

01:02 And those are the x components of those two forces.

01:05 And so those have to balance in order for point c to be at equilibrium.

01:10 And for the future, i'm just going to go ahead and solve this for t .a in terms of tb by dividing both sides by the cosine of alpha.

01:19 And then we're going to use this in just a minute.

01:23 Now looking at the y direction, the forces acting in the wide direction, both the tension in ac and the tension in b, b, c are going to be pointed up.

01:35 And so the y components of those combined, so t .a.

01:38 Sine alpha plus tb sine beta, will equal the downward force of force p.

01:45 So p cosine theta.

01:48 We can then substitute what we just found for ta into this equation, and that way we'll have one equation with one unknown.

01:56 So tb cosine beta times the sign of alpha divided by cosine alpha becomes tangent alpha.

02:04 So tb cosine beta tangent alpha.

02:07 And then i'm just distributing the sign alpha through that ta term...

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