11.66 A holiday decoration consists of two shiny glass spheres with masses 0.024

Question

11.66 A holiday decoration consists of two shiny glass spheres with masses 0.0240 kg and 0.0360 kg suspended, as shown in the figure (Figure 1) , from a uniform rod with mass 0.120 kg and length 1.00 m. The rod is suspended from the ceiling by a vertical cord at each end, so that it is horizontal.

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A:Calculate the tension in the cord A .

B:Calculate the tension in the cord B .

C:Calculate the tension in the cord C .

D:Calculate the tension in the cord D

E:Calculate the tension in the cord E

F:Calculate the tension in the cord F .

Explanation / Answer

Starting with the lower mass (.036), the only forces acting on it are the tension in cord A (Ta) and it's weight (mg) so

Ta = mg = 0.036(kg) * 9.8 (m/s^2) = 0.353 N

For the other mass, the forces are Ta (down), Tb (up), and weight (down)

Tb = Ta + mg = 0.353 + 0.024*9.8 = 0.588 N

Consider the point where B, C, and D intersect. The forces acting on this point must sum to zero.
Horizontal:
Tc * cos(53.1) = Td * cos(36.9)
and Vertical:
Tc * sin(53.1) + Td * sin(36.9) = Tb ...(1)

so,
Tc = Td * (cos(36.9)/cos(53.1)) ...(2)
substitute this in (1) and solve for Td

Td = Tb / [sin(36.9) + sin(53.1)*cos(36.9)/cos(53.1)] = 0.353 N

using this value of Td in (2):
Tc = 0.470 N

To get Te and Tf, pick a couple of points on the rod and sum the torques. They must add up to zero. If you choose the point where cord E joins the rod...
(clockwise torque = counterclockwise torque)
(torque due to D + torque due to C + torque due to mass of rod) = torque due to Tf
0.2*Td*sin(36.9) + 0.8*Tc*sin(53.1) + 0.5*0.120*9.8 = 1.0*Tf
Tf = 0.931N

and where F joins the rod...
1.0*Te = 0.2*Tc*sin(53.1) + 0.8*Td*sin(36.9) + 0.5*0.120*9.8
Te = 0.833N

Note that the sum of the tensions in E and F should equal the sum of the three weights:
0.931 + 0.833 = 9.8 * (0.036 + 0.024 + 0.120)
1.764 = 1.764

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