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# calculate x and y components?

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 horizontal-right. The second string ( T2 )

makes an angle of 20 degree with the horizontal-left. The

equations relating the x-components and the y-components

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. :(

### 3 Answers

- NCSLv 75 years agoFavorite Answer
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 horizontal-right. The second string ( T2 ) makes an angle of 20 degree with the horizontal-left. The equations relating the x-components and the y-components 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.

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- 5 years ago
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.

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- electron1Lv 75 years ago
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 horizontal-right and the second string is at an angle of 20 degree with the horizontal-left. 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.

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