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It is the most powerful prayer. A pure heart, a clean mind, and a clear conscience is necessary for it.
- Samuel Dominic Chukwuemeka

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# Word Problems on Vectors

Calculators: Calculators

Prerequisites:
(1.) Expressions and Equations
(2.) Trigonometry

For ACT Students
The ACT is a timed exam...60 questions for 60 minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for a wrong answer.

For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.
Unless specified otherwise, any question labeled JAMB is a question from JAMB Physics

For WASSCE Students: Unless specified otherwise:
Any question labeled WASCCE is a question from WASCCE Physics
Any question labeled WASSCE-FM is a question from the WASSCE Further Mathematics/Elective Mathematics

For NYSED Students
Unless specified otherwise, any question labeled NYSED is a question from NYSED Physics

For GCSE Students
All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve all questions.
Show all work.

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(2.) ACT (Science) If an object is at rest and the sum of the forces acting on it is zero, the object is in a state of static equilibrium.
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:
$55.6\hat{x} + 0.0\hat{y}$
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