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A thin uniform rod of mass M is suspended horizontally by two vertical wires. One wire is at the left end of the rod, and the other wire is 3/4 of the length of the rod from the left end.

(a) T = ?

Determine the tension (T) in each wire. (Use g for the acceleration due to gravity and M as necessary.)

(b) M = ?

An object is now hung by a string attached to the right end of the rod. When this happens, it is noticed that the rod remains horizontal but the tension in the wire on the left vanishes. Determine the mass m of the object. (Use M for the mass of the rod as necessary.)

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Edited Sun, 18 Oct '15
Added Fri, 05 Jun '15
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What are our Givens?

M = mass

L = length of rod

d1= L

d1 = distance from the left end of wire 1

d2 = 1/4L

d1 = distance from the left end of wire 2

G = gravity

(a) Determine tension in each wire using M for mass and g for gravity.

Because the system in in equilibrium - the vertical forces, horizontal forces, and torque must all have a net of 0.

Thus:

Newton's Third Law in the Y direction gives us :

Fy = T_1 + T_2 - Mg = 0

>> T_1 + T_2 = Mg

Definition of Torque (or Moments), with the point of reference at the left end of the rod, gives us:

Tq = L/2*Mg - d1*T_1 - d2*T_2 = 0

>> L/2*Mg - 0*T_1 - 3/4L * T_2 = 0

>>1/2 * Mg - 3/4 * T_2 = 0 (divide by L)

>> 1/2 * Mg = 3/4 T_2

>> 2/3 Mg = T_2

Solve for T_1

We know T_1 + T_2 = Mg

We know T_2 = 2/3Mg

Thus:

T_1 + 2/3Mg = Mg

T_1 = 1/3 Mg

(b) an object is attached to the right side of the rod, when this happens the tension in the left becomes 0. What's the mass of this object? (m)

Repeat the process used in (a) to solve for m.

Newton's Third Law in the Y direction gives us :

Fy = T_2 - Mg - mg = 0

>> T_2 = Mg + mg

Definition of Torque (or Moments), with the point of reference at the Right end of the rod, gives us:

Tq = 0*mg - L/2*Mg + 1/4L * T_2

1/4L * T_2 = L/2 * Mg

T_2 = 2Mg (divide by 1/4L)

Solve for mg

We know T_2 = Mg + mg

We know T_2 = 2Mg

Thus:

2Mg = Mg + mg

Mg = mg

M = m

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Added Fri, 05 Jun '15
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