Find the tension in each cord in Fig. E5.7 if the weight of the suspended object is w. Figure e5

Find the tension in each cord in Fig. E5.7 if the weight of...

Key Concepts

Symmetry in Equilibrium Systems

When an object is suspended by cords that are arranged symmetrically, the system's inherent symmetry simplifies the analysis. The symmetry can often lead to the conclusion that the tensions in the cords are equal, or at least simplifies the relationship between them, by reducing the number of independent variables in the equilibrium equations.

Force Decomposition

This concept involves breaking down forces into their component parts, typically along perpendicular axes (often horizontal and vertical). By resolving the tension forces into vertical and horizontal components, one can set up equilibrium equations to find the unknown tensions, especially when the cords are oriented at angles.

Tension in Cords

Tension refers to the force conducted along a rope, string, or cord when forces are applied at its ends. In suspended systems, particularly those with multiple cords, determining the tension in each cord involves ensuring that the net forces in all directions are balanced.

Static Equilibrium

This is the state in which an object remains at rest because the sum of forces and the sum of moments acting on it are zero. In problems involving suspended objects, applying the condition for static equilibrium ensures that the upward forces (tensions) balance the downward force (weight) of the object.

Transcript

00:01 Okay, in this problem we're given two different scenarios where we have a weight hanging off of some cords and you're asked to calculate the tension in each one of the cords here.

00:14 So let's draw out our situation.

00:16 We have our first diagram here and we have our weight hanging off and we have c, a, and b.

00:32 And the a chord is at a 30 degree angle, and the b chord here is at a 45 degree angle to the ceiling.

00:44 So i mark points one and two here, and we're gonna draw the forced body diagrams for each point.

00:53 So the first body, the force body diagram for one is pretty simple.

00:57 You just have chord c, the tension in that pointed upwards, and you'd have the weight of our object pointed downwards.

01:08 And our second force body diagram, we have a few different forces pointed in different directions.

01:15 So we have a pointed off like so, b directed upwards, and now c is pointed directly below.

01:31 And just so we have thrown our reference frame here.

01:38 So i might break down a and b here into the components of c, just so we stay within the x, y reference frame that i've established.

01:48 So we see that c will just stay the same.

01:56 And if we break the b out in the components, the vertical component of the tension due to chord b will be b sign of 45 degrees and the horizontal component then is b cosine of 45 degrees and we're going to do the same thing with chord a.

02:26 You see the vertical component is going to be a times the sign of 30 degrees and the horizontal component is a times the cosine of 30 degrees.

02:53 Okay, so we have a force body diagrams.

02:56 We have three unknowns in our tensions in the three different chords and let's see we're gonna have three equations we're gonna set up here.

03:07 So from free body diagram number one we just have the tension due to c minus the weight is equal to zero assuming no acceleration.

03:17 So we see that c is just equal to to w.

03:23 In our second part, let's do the horizontal components first.

03:29 We have b times the cosine of 45 degrees minus a cosine of 30 degrees equals 0.

03:51 And here we have b, sign of 45 degrees plus a sign of 30 degrees minus c is equal to zero.

04:25 So first things first, i see i can substitute w in for c here and move it over to the other side.

04:31 And that will give me this equation here.

04:39 And then now i have a system of equation with two unknowns and two equations here and here.

04:57 So i want to eliminate one of my unknowns here and then solve for the other.

05:03 So we can use our knowledge of trig to say that cosine of 45 degrees is equal to sign of 45 degrees.

05:14 So if we just subtract these two equations from each other, the first terms in each one will cancel each other out.

05:23 So that would leave us with negative a cosine of 30 degrees.

05:36 Plus sign of 30 degrees is equal to negative w.

05:47 So those negative signs on either side cancel out.

05:52 And in the end, you get that a is equal to w over cosine of 30 degrees plus the sign.

06:11 Of 30 degrees.

06:16 And when you plug in the numbers for cosine of 30 and sign of 30 degrees, you'd get that a is equal to 0 .732w.

06:37 Now let's move over here to now solve for b.

06:42 So from this equation we can get b in terms of a.

06:49 So it will just be a times the cosine of 30 degrees times the cosine of 45 degrees...

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