Determine the tension developed in each cord required for equilibrium of the 20-kg lamp.
Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.
Solution:
Show me the final answers↓
We will start off by drawing a free body diagram focusing on ring D.
Let us write an equation of equilibrium for the y-axis forces.
F_{DE}\text{sin}\,(30^0)\,-\,196.2\,=\,0
(Solve for F_{DE})
F_{DE}\,=\,392.4 N
Now, we will write an equation of equilibrium for x-axis forces.
F_{DE}\text{cos}\,(30^0)\,-\,F_{DC}\,=\,0
(Substitute the value of F_{DE} we found)
392.4\text{cos}\,(30^0)\,-\,F_{DC}\,=\,0(solve for F_{DC})
F_{DC}\,=\,339.8 N
We can now draw a free body diagram focusing on ring C.
Again, we will write our equations of equilibrium.
339.8\,-\,F_{CA}\dfrac{3}{5}\,-\,F_{CB}\text{cos}\,(45^0)\,=\,0 (eq.1)
(Remember that we already found F_{DC}\,=\,339.8 N)
+\uparrow \sum \text{F}_\text{y}\,=\,0
F_{CA}\dfrac{4}{5}\,-\,F_{CB}\text{sin}\,(45^0)\,=\,0 (eq.2)
Let us solve for F_{CA} and F_{CB}
(Simplify)
F_{CA}\,=\,0.88F_{CB} (eq.3)
Substitute this value into eq.1.
(solve for F_{CB})
F_{CB}\,=\,275.1 N
Substitute this value back into eq.3 to figure out F_{CA}.
F_{CA}\,=\,242.1 N
Final Answers:
F_{DC}\,=\,339.8 N
F_{CB}\,=\,275.1 N
F_{CA}\,=\,242.1 N
where did -196.2 came when equation of equlibruium for the y-axis forces ?
Hi!
The 196.2 N comes from 20 kg multiplied by the acceleration due to gravity, 9.81. Thus, we get 20 kg x 9.81 m/s = 192.6 N. Also, note that this is shown in the first force diagram, with the light blue arrow pointing downwards.
Hope that helps 🙂
HOW ABOUT F,df?
F is just mass times the acceleration due to gravity, that’s also the tension in the cord.
what if we did not have the triangle of 3 and 4 and 5 on AC how would you solve it
Something else must be given, like an angle, or the lengths to figure out the angle. Maybe the question already gives the tension in AC? I am not sure unless I see the question. Please see: https://www.youtube.com/watch?v=X9g4G1eBHCA
sorry can u explain how we got From
CB=275.1 N
So you’re just solving for a single variable. If it’s easier, convert the cosine values into decimals, so cos(45) = 0.707. That should be easier to visualize.