# Finding the Tension

1. Jun 25, 2015

### DameLight

Solved Thank You : )
1. The problem statement, all variables and given/known data

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.

2. Relevant equations
Στ = Iα = F × r
Ibar = 1/12 m1L2

3. 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
TA = [m1g(L/2) + m2gx]/d

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

Last edited: Jun 25, 2015
2. Jun 26, 2015

### theodoros.mihos

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 $\tau_{m_2}+\tau_{bar}$.