Tension in the cables.?

Question

Ok I have a block of 200lbs hanging from two strings located at Two points. One is two units to the left of the center of the ring holding the block and 3 units up from the block. The other string is attached one unit to the right and 2 units up. Should the magnitude of the left vector be more then 200lbs? Is the concept here that my left vector plus my right vector = my downward vector?

Answer 1

The vertical components of the strings will be equal to 200 lbs. However, if the strings are not vertical, there will be horizontal components in them.

If I am understanding the scenario, the left string is pointing upwards at an angle of arc-tan (3/2) (about 53 deg. from horizontal) and the right string is at an angle of arc-tan (2/1) (about 63 deg. from horizontal). Use those angle to determine the vertical and horizontal components of each string. The sum of the 2 vertical components will be 200 lbs.

Hope this helps.

Answer 2

The concept is that the vertical component of the left vector plus the vertical component of the right vector should be equal in magnitude (and opposite in direction) to the downward vector.

Similarly, the vector sum of the horizontal component of the left and right vectors should equate to zero.

Answer 3

Before you start, draw a free body diagram of the situation. Making the origin at the center of the block. I labeled the string 2 units to the left and 3 units up T1, and the other one T2. Don't forget to put in the angles for T1 and T2, which will be found by doing arctan(3/2)=(56.31) and for T2 it will be 60 because of the 30-60-90 triangle rule. Also, don't forget to convert the 200pounds to Newtons. The conversion is 4.448N=1pound, so 200pounds=889.6N.

Now you are ready to sum the forces in the X and Y.
Fx=-T1cos(56.31)+T2cos(60)=0
Fy=T2sin(60)+T1sin(56.31)-889.6N=0

These sums equal zero because of Newton's second law, which states that the sum of the forces is equal to mass*acceleration, and since the structure is not moving, the acceleration is 0.

Now, if you've taking algebra, you know that if you have two equations and two unknowns, you can solve for one and plug it into the other and solve.

Solve for T2 in the Fx equation and get
T2=(T1cos(56.31))/cos(60)
plug T2 into Fy and get
T1cos(56.31)*tan(60)+T1sin(56.31)-889.6=0
Factor out T1 and move 889.6 to the other side and get
T1(cos(56.31)*tan(60)+sin(56.31))=889.6N
Now solve for T1 and get
T1=889.6/(cos(56.31)*tan(60)+sin(56.31))
T1=+496.202N
Plug T1 into Fx and solve for T2
-(496.202)cos(56.31)+T2cos(60)=0
T2=(496.202)cos(56.31)/cos(60)
T2=+550.49N

Now that you have T1 and T2, +496.202N and +550.49N you can determine if the strings are in tension or compression. The strings are in compression if the internal forces are negative, and they are in tension if the internal forces are positive. Since the the Forces are both positive, both of the strings are in Tension.

So, the answer to your question is Yes, the strings are in Tension.