A 3.45 kg object is attached to a vertical rod by two strings as in Figure P6.11. The object rotates in a horizontal circle at constant speed 5.65 m/s. What are the tensions of the two strings, individually?(Figure: http://www.webassign.net/pse/p6-11.gif)
I assumed that the ball was half way between, thus we have a triangle with hypotenuse of 2m and side 1.5m and by solving we find the radius to = sqrt(1.75) = 1.32m. We can find the sum of the forces by using m(v^2/r): (3.45kg)((5.65m/s)^2/1.32m) = 83.25N. I am lost how to proceed assuming all my prior work is correct? Also, can I assume that both strings have the same tension in this situation? Any help would be appreciated, thanks!
I assumed that the ball was half way between, thus we have a triangle with hypotenuse of 2m and side 1.5m and by solving we find the radius to = sqrt(1.75) = 1.32m. We can find the sum of the forces by using m(v^2/r): (3.45kg)((5.65m/s)^2/1.32m) = 83.25N. I am lost how to proceed assuming all my prior work is correct? Also, can I assume that both strings have the same tension in this situation? Any help would be appreciated, thanks!
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3.45 kg object is attached to a vertical rod by two strings as in Figure P6.11. The object rotates in a horizontal circle at constant speed 5.65 m/s.
The horizontal distance from the rod to the object = (2^2 – 1.5^2)^0.5 =
√1.75
The force which is keeping the object moving in a circle is the centripetal force = m * v^2/r
r is the radius = horizontal distance from the rod to the object = √1.75 m.
centripetal force = (3.45 * 5.65^2) ÷ (√1.75) = 83.25 N
The 2 strings must exert enough force to make a total of 83.25 N horizontally
The horizontal component of the tension of each string = ½ * 83.25 = 41.625 N
The horizontal component of the tension = T * cos θ
The strings are at an angle from horizontal.
The cosine of the angle = (horizontal distance) ÷ (length of string)
The cosine of the angle = (√1.75)/2 = 0.6614
The angle = Inverse tan (0.6614) = 48.6˚
This is the angle between horizontal and each string.
T * cos 48.6 = 41.625 N
Tension = 41.625 ÷ cos 48.6 = 62.94 N
Check:
With no rounding:
Tension = (horizontal component of the tension) ÷ (Cos θ)
The horizontal component of the tension = ½ * (3.45 * 5.65^2) ÷ (√1.75)
Cos θ = (√1.75/2)
Tension = [½ * (3.45 * 5.65^2) ÷ (√1.75)] ÷ (√1.75/2) = 62.93 N
This is one complex problem!
I hope you can follow all the work above.
The horizontal distance from the rod to the object = (2^2 – 1.5^2)^0.5 =
√1.75
The force which is keeping the object moving in a circle is the centripetal force = m * v^2/r
r is the radius = horizontal distance from the rod to the object = √1.75 m.
centripetal force = (3.45 * 5.65^2) ÷ (√1.75) = 83.25 N
The 2 strings must exert enough force to make a total of 83.25 N horizontally
The horizontal component of the tension of each string = ½ * 83.25 = 41.625 N
The horizontal component of the tension = T * cos θ
The strings are at an angle from horizontal.
The cosine of the angle = (horizontal distance) ÷ (length of string)
The cosine of the angle = (√1.75)/2 = 0.6614
The angle = Inverse tan (0.6614) = 48.6˚
This is the angle between horizontal and each string.
T * cos 48.6 = 41.625 N
Tension = 41.625 ÷ cos 48.6 = 62.94 N
Check:
With no rounding:
Tension = (horizontal component of the tension) ÷ (Cos θ)
The horizontal component of the tension = ½ * (3.45 * 5.65^2) ÷ (√1.75)
Cos θ = (√1.75/2)
Tension = [½ * (3.45 * 5.65^2) ÷ (√1.75)] ÷ (√1.75/2) = 62.93 N
This is one complex problem!
I hope you can follow all the work above.