Yahoo Answers: Answers and Comments for Calculate x and y components? [Physics]
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From Eren
enUS
Sun, 28 Dec 2014 18:14:03 +0000
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Yahoo Answers: Answers and Comments for Calculate x and y components? [Physics]
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https://answers.yahoo.com/question/index?qid=20141228181403AAzrgNE
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From Bob: Draw a free body diagram for the object. You s...
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https://answers.yahoo.com/question/index?qid=20141228181403AAzrgNE
Sun, 28 Dec 2014 18:27:23 +0000
Draw a free body diagram for the object. You should have 1 arrow pointing down, which is the weight of the object. Then, you should have 2 arrows at different angles to the left and right respectively pointing upward from the object, which represent the tensions.
Since the mass isn't moving, the overall net force is 0
Now split the 2 tensions into their vertical and horizontal components. Let's call the 20° tension T1 and the 50° one T2
Since the object is not moving, both the vertical and horizontal directions must equate to 0, respectively.
For the horizontal direction the 2 horizontal components of the tensions are the acting forces and must equate to 0 meaning that the tensions are acting against each other, giving us
T2cos50  T1cos20 = 0
*This also means that the horizontal components of the tensions must be equal to each other* (T2cos50 = T1cos20)
For the vertical direction the forces acting are the weight of the object and both of the vertical components of the tensions.Since the object is not moving in the vertical direction either, the forces must once again equate to 0. Also remember that the weight is acting downwards while BOTH of the vertical components are acting upwards, so the components of the tensions must be added. Finally, this gives us (remember weight = mass * gravity)
Weight = (900 kg)(9.8 m/s²) = 8820 N
T1sin20 + T2sin50  8820 N= 0
*This also means that the two vertical components of the tensions must equal the weight of the object*(T1sin20 + T2sin50 = 8820 N)
Hope this helps.

From NCS: Q: "An object of mass 900 kg is hanging f...
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Mon, 29 Dec 2014 20:03:13 +0000
Q: "An object of mass 900 kg is hanging from a ceiling by means of two strings. The first string ( T1 ) makes an angle of 50 degree with the horizontalright. The second string ( T2 ) makes an angle of 20 degree with the horizontalleft. The equations relating the xcomponents and the ycomponents of the forces acting on the object respectively are
answer: 0.643 * T1 + 0.94 * T2 = 0
0.766 * T1 + 0.342 * T2  8820 N = 0
i understand how to get the .766 and .643 but I dont understand why .643 is negative. besides that, I dont understand any of it. :( "
A: You didn't say which equation was which!
Let's say "right" and "up" are positive.
For equilibrium, the net forces horizontally and vertically must be zero.
Horizontally, we have
Σ F = 0 = T1 * cos50  T2 * cos20 = 0.643*T1  0.940*T2
As you can see, I have the signs reversed from what you have. This is fine, since the LHS is zero; I've just multiplied my equation by 1. (Or, to put it another way, you've taken "left" as positive.) Whatever the convention, it must be clear that the two horizontal components act in opposite directions, so they must have different signs. So
T2 = 0.684*T1
Vertically we have
Σ F = 0 = T1*sin50º + T2*sin20º  900kg * 9.8m/s²
8820 N = 0.766*T1 + 0.342*T2
The vertical components act in the same direction (up) and should have the same sign, opposite that of the weight.
Substituting for T2, we have
8820 N = 0.766*T1 + 0.342*0.684*T1 = 1.00 * T1
→ T1 = 8820 N ◄ tension in right cable
→ T2 = 0.684*T1 = 0.684 * 8820N = 6030 N ◄ tension in left cable
after rounding to three significant digits.
In short, for problems like this you need to equate the horizontal components so that you can express one tension as a function of the other. Then you need to equate the vertical components to the weight, substitute for one of the tensions, and solve for the other tension. You then solve for the second tension, as above.
If you find this helpful, please select Best Answer!

From electron1: In the first the negative sign on 0.643 * T1 m...
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Sun, 28 Dec 2014 19:50:07 +0000
In the first the negative sign on 0.643 * T1 means direct of this x component the opposite of the direction of the 0.94 * T2. This is because the first string is at an angle of 50 degree with the horizontalright and the second string is at an angle of 20 degree with the horizontalleft. To prevent the object from moving left or right, the magnitudes of the horizontal components must be equal.
T1 * cos 50 = T2 * cos 20
T2 = T1 * cos 50 ÷ cos 20
The sum of the vertical components of the two tension forces is equal to the object’s weight.
Vertical component = T * sin θ, Weight = 900 * 9.8 = 8820 N
T1 * sin 50 + T2 * sin 20 = 8820
This is the second equation that you are given. Let’s substitute (T1 * cos 50 ÷ cos 20) for T2 in the equation above and solve for T1.
T1 * sin 50 + (T1 * cos 50 ÷ cos 20) * sin 20 = 8820
T1 * sin 50 + T1 * cos 50 * tan 20 = 8820
T1 * (sin 50 + cos 50 * tan 20) = 8820
(sin 50 + cos 50 * tan 20) = 1
T1 = 8820 N
Let’s use this number in the following equations to determine T2.
T2 = T1 * cos 50 ÷ cos 20
T2 = 8820 * cos 50 ÷ cos 20
This is approximately 6033 N.
T1 * sin 50 + T2 * sin 20 = 8820
8820 * sin 50 + T2 * sin 20 = 8820
T2 * sin 20 = 8820 – 8820 * sin 50
T2 = (8820 – 8820 * sin 50) ÷ sin 20
This is approximately 6033 N.
Since both of these answers are the same, I believe that T2 is correct. I hope this helps you to understand how to solve this problem.