(a) Calculate the tension in the cable between the rod and the wall, assuming the cable is holding the system in the position shown in the figure. (b) Find the horizontal force exerted on the base of the rod. (c) Find the vertical force exerted on the base of the rod. Ignore the weight of the rod.

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12.4

A 8 050-N shark is supported by a rope attached to a 3.10-m rod that can pivot at the base.
20.0°
60.0⁰
(a) Calculate the tension in the cable between the rod and the wall, assuming the cable is holding the
system in the position shown in the figure.
(b) Find the horizontal force exerted on the base of the rod.
(c) Find the vertical force exerted on the base of the rod. Ignore the weight of the rod.
Transcribed Image Text:A 8 050-N shark is supported by a rope attached to a 3.10-m rod that can pivot at the base. 20.0° 60.0⁰ (a) Calculate the tension in the cable between the rod and the wall, assuming the cable is holding the system in the position shown in the figure. (b) Find the horizontal force exerted on the base of the rod. (c) Find the vertical force exerted on the base of the rod. Ignore the weight of the rod.
(a) From the diagram, the angle T makes with the rod is given by
0 = 60
° +20.0° = 80
and the perpendicular component of Tis
T sin 80
Summing torques around the base of the rod, and applying Newton's second law in the horizontal and vertical
directions, Σ T = 0, and we have
- (³ 3.1
m)(
sin 80° = 0,
which gives
T =
8050
4.08709211284
=
(b) From ΣΕx 0, we have FH
X
=
8050
N) cos(60
sin(80.0⁰)
x 10³ N.
os(20. .0°) =
= T cos 20.0°
✓
Ncos 60
= 0, so
O
+
3.1
m
FH = T cos(20.0⁰) = = 4.087
Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a
mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize
roundoff error. × 10³ N.
Transcribed Image Text:(a) From the diagram, the angle T makes with the rod is given by 0 = 60 ° +20.0° = 80 and the perpendicular component of Tis T sin 80 Summing torques around the base of the rod, and applying Newton's second law in the horizontal and vertical directions, Σ T = 0, and we have - (³ 3.1 m)( sin 80° = 0, which gives T = 8050 4.08709211284 = (b) From ΣΕx 0, we have FH X = 8050 N) cos(60 sin(80.0⁰) x 10³ N. os(20. .0°) = = T cos 20.0° ✓ Ncos 60 = 0, so O + 3.1 m FH = T cos(20.0⁰) = = 4.087 Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. × 10³ N.
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