A nonuniform, horizontal bar of mass m is supported by two massless wires agains

Question

A nonuniform, horizontal bar of mass m is supported by two massless wires against gravity. The left wire makes an angle Phi_1 with the horizontal, and the right wire makes an angle Phi_2. The bar has length L. Find the position of the center of mass of the bar x, measured from the bar's left end Express the center of mass in terms of L, Phi_1, and Phi_2. There is an error in your submission Make sure you have formatted it properly and any variable subscripts are added immediately following the variable

Explanation / Answer

So, write all the equations for balancing the forces assume tension in wire with phi1 is T1, and tension in wire with phi2 angle is T2.

So, T1cos(phi1) = T2cos(phi2)
and T1sin(phi1) + T2sin(phi2) = mg
using these two equations obtain T1 and T2

T1 = (mgcos(phi2))/(sin(phi1+phi2)) (Note that sin(phi1+phi2) = sinphi1cosphi2+sinphi2cosphi1)
T2 = (mgcos(phi1))/(sin(phi1+phi2))

now balance the moments about the centre of mass which is x distances from left edge of the bar.
Then T1(sinphi1)*x = T2sin(phi2)*(L-x)
x = (T2*L*sin(phi2))/(T1sin(phi1)+T2sin(phi2))
Substitute T1 and T2
So, x = L* (cos(phi1)sin(phi2))/(cos(phi2)sin(phi1) + cos(phi1)sin(phi2))

If we divide this by cosphi1*cosphi2 then we obtain the same result which the system has denied to accept try using alternate forms as answer and it may accept them.

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