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

Question

A nonuniform horizontal bar of mass m is supported by two massless wires against gravity. The left wire makes an angle ?1 with the horizontal, and the right wire makes an angle ?2. The bar has length

A nonuniform horizontal bar of mass m is supported by two massless wires against gravity. The left wire makes an angle ?1 with the horizontal, and the right wire makes an angle ?2. The bar has length What is the position of the center of mass of the bar, measured as distance x from the bar's left end? Express your answer in terms of ?1, ?2, and L.

Explanation / Answer

Let the tensions in the ropes be T1 N on left and T2 N on right side.

now writing the equations for rotational and translational equilibria

for translational we get

T1 sin ?1 + T2 sin ?2 = mg (where m is the mass of rod) and

T1 cos ?1 = T2 cos ?2

that is T1 = T2 (cos ?2 /cos ?1 ) - (1)

for rotational ( if we assume that center of mass is x units away from left end)

( T1 sin ?1) * x = (T2 sin ?2) * ( L - x )

this can be re-written as

( T1 sin ?1 + T2 sin ?2 ) * x = (T2 sin ?2) * L

substitute the value of T1 from (1)

T2[ (cos ?2 /cos ?1) * sin ?1 )+ sin ?2 ]* x = (T2 sin ?2) * L

now solving for x , we get

(cos ?2 /cos ?1) * sin ?1 )+ sin ?2 ]* x = ( sin ?2) * L

taking out cos ?2 as common on left side and re-arranging

[ tan ?1 + tan ?2 ] * x = ( tan ?2) * L

now solving for x we get

x = [( tan ?2) * L ] / [ tan ?1 + tan ?2 ]

that can be re-arranged as

x = L / [( tan ?1 / tan ?2 ) + 1]   

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