Finding the Tension

  • Thread starter DameLight
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Solved Thank You : )
1. Homework Statement

A rigid, uniform, horizontal bar of mass m1 and length L is supported by two identical massless strings. (Figure 1) Both strings are vertical. String A is attached at a distance d < L/2 from the left end of the bar and is connected to the ceiling; string B is attached to the left end of the bar and is connected to the floor. A small block of mass m2 is supported against gravity by the bar at a distance x from the left end of the bar, as shown in the figure.

Throughout this problem positive torque is that which spins an object counterclockwise. Use g for the magnitude of the acceleration due to gravity.

m1
L
dA < L/2
m2
dm2 = x
g

Find TA, the tension in string A.

Homework Equations


Στ = Iα = F × r
Ibar = 1/12 m1L2

The Attempt at a Solution


the rod isn't moving so
Στ = 0

If I make the point of rotation the location of string B
Στ = 0 = - τbar - τm2 + τA

τbar = Fg × L/2 = m1g(L/2)
τm2 = Fg × x = m2gx

τA = T × dA = TdA(sin90)
= τbar + τm2
= m1g(L/2) + m2gx

dA = this is a known variable so
TAd = m1g(L/2) + m2gx
TA = [m1g(L/2) + m2gx]/d

Conclusion:
Reason for error- the answer did not require the cross product to have sinθ
 
Last edited:

Answers and Replies

  • #2
All forces are 90 degrees cross to lengths. ##\sin(90)=1## for all. Your calculation is correct. The rope of T according to B point must be opposite to total masses rope ##\tau_{m_2}+\tau_{bar}##.
 

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