Word Problems on Vectors
Calculators: Calculators
Prerequisites:
(1.) Expressions and Equations
(2.) Trigonometry
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For the Questions:
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Show all work.
In each trial of 2 experiments on static equilibrium, 3 cables and a cylinder were arranged as shown in Figure 1.
Cables 1, 2, and 3, each with it own tensiometer (a device used to measure tension), were joined with a single knot.
The free ends of Cables 1 and 2 were attached to an L-shaped support, and a cylinder of mass M was suspended from the free end of Cable 3.
Cable 1 made an angle θ with the top of the support, and Cable 2 made a 90° angle with the side of the support.
None of the cables stretched under tension.
The tension forces acting on the knot, which was in static equilibrium, are shown in Figure 2.
The tension in Cable 1, $T_1$, had a horizontal component, $T_{1x}$, and a vertical component, $T_{1y}$
The tension in Cable 2, $T_2$, was purely horizontal, and the tension in Cable 3, $T_3$, was purely vertical.
Experiment 1
In Trials 1-5, M was 1.0 kg and θ was varied.
Table 1 lists θ, in degrees, as well as $T_{1x}$, $T_{1y}$, $T_2$, and $T_3$, each in newtons (N), for each trial.
| Table 1 | |||||
| Trial | θ |
$T_{1x}$ (N) |
$T_{1y}$ (N) |
$T_2$ (N) |
$T_3$ (N) |
|
1 2 3 4 5 |
10° 30° 50 ° 70 ° 90 ° |
-55.6 -17.0 -8.2 -3.6 0.0 |
9.8 9.8 9.8 9.8 9.8 |
55.6 17.0 8.2 3.6 0.0 |
-9.8 -9.8 -9.8 -9.8 -9.8 |
Experiment 2
In Trials 6-10, M was varied and θ was 50°.
Table 2 lists M as well as $T_{1x}$, $T_{1y}$, $T_2$, and $T_3$, for each trial.
| Table 2 | |||||
| Trial |
M (kg) |
$T_{1x}$ (N) |
$T_{1y}$ (N) |
$T_2$ (N) |
$T_3$ (N) |
|
6 7 8 9 10 |
2.0 3.0 4.0 5.0 6.0 |
-16.4 -24.7 -32.9 -41.1 -49.3 |
19.6 29.4 39.2 49.0 58.8 |
16.4 24.7 32.9 41.1 49.3 |
-19.6 -29.4 -39.2 -49.0 -58.8 |
A vector quantity, such as tension, may be written as $A\hat{x} + B\hat{y}$, where A is the vector's horizontal component and B is the vector's vertical component.
For example, the tension (in N) in Cable 1 during Trial 1 could be written as $-55.6\hat{x} + 9.8\hat{y}$
Which of the following expressions gives the tension (in N) in Cable 2 during Trial 1?
$ A.\;\; -55.6\hat{x} - 9.8\hat{y} \\[3ex] B.\;\; -55.6\hat{x} + 0.0\hat{y} \\[3ex] C.\;\; 55.6\hat{x} - 9.8\hat{y} \\[3ex] D.\;\; 55.6\hat{x} + 0.0\hat{y} \\[3ex] $
During Trial 1:
The tension in Cable 2, $T_2$, was purely horizontal
horizontal component is the x-component
This implies that $T_2 = T_{2x} = 55.6\hat{x}$
This also implies that $T_{2y} = 0\hat{y}$
Hence, the vector quantity for the tension (in N) in Cable 2 during Trial 1 is:
$55.6\hat{x} + 0.0\hat{y}$