Question

Please please please help

10 A river is flowing at a speed of 1.5 m/s. Sam wants to row from point A on one
bank to point B on the other bank directly opposite A. He intends to maintain a
constant speed of 2.5 m/s. In what direction, correct to the nearest degree, should Sam row? Give your
answer as an angle of inclination to the line AB.
11 Two dogs Brutus and Nitro are simultaneously tugging on a bone. Brutus is pulling with a force of 12 N
in a direction 45° west of north, while Nitro is pulling with a force of 16 N in a direction 30° south of
east. Calculate, correct to two significant figures, the magnitude and direction of the resultant force.
12 Three forces act on an object of mass 5 kg. These forces are represented by the vectors 9į - 2j,
-3i+ 10j and 18i - j. Calculate the magnitude and direction of the acceleration of the object.
7 The diagram shows an object of weight 24 N at rest
at P on an inclined plane. Find the component of
the weight:
a down the plane,
b
perpendicular to the plane.
30°
24N
30°
8 In the diagram, the vectors OP and Og represent forces of magnitude 20 N and
16 N respectively.
a Express OP and O0 in component form.
b
Calculate, correct to 2 significant figures, the magnitude and direction of the
resultant of the two forces.

Answer 1

Sam should row at an angle of approximately 30 degrees to the line AB.

The magnitude of the resultant force is approximately 19.5 N and it is in a direction of approximately 37 degrees south of west.

The magnitude of the acceleration of the object is approximately 5.56 m/s^2 and its direction is approximately 12 degrees east of north.

The component of the weight down the plane is 12 N and the component perpendicular to the plane is 20.8 N.

8a. OP can be expressed as 20cos(45)i + 20sin(45)j, and O0 can be expressed as -16i.

8b. The magnitude of the resultant force is approximately 8.2 N and it is in a direction of approximately 70 degrees above the negative x-axis.

Answer 2

10. To row directly across a river, Sam should row at an angle of approximately 30.96° to the line AB.

11. The resultant force on the bone is approximately 22.61 N at 105° south of west.

12. The object's acceleration is approximately 5.07 m/s² at 16.26° north of east.

7. Component down the plane is 12 N, and component perpendicular to the plane is approximately 20.78 N.

8. Resultant force magnitude is calculated as approximately 24.77 N, and direction is found as 45.96° south of east.

Sure, let's tackle each problem step by step:

Problem 10:

Sam wants to row from point A to point B across a river. The velocity of the river is 1.5 m/s, and Sam's rowing speed is 2.5 m/s. To find the direction he should row, we use vector addition.

Let \(\theta\) be the angle between Sam's rowing direction and the river bank.

`\[ \tan(\theta) = \frac{\text{velocity of the river}}{\text{rowing speed}} \]`

`\[ \tan(\theta) = (1.5)/(2.5) \]`

`\[ \theta = \tan^(-1)(0.6) \]`

`\[ \theta \approx 30.96° \]`

So, Sam should row at an angle of approximately
`\(30.96°\)` to the line AB.

Problem 11:

Given two forces, Brutus (12 N at 45° west of north) and Nitro (16 N at 30° south of east), we can find the resultant force using vector addition.

`\[ F_{\text{resultant}} = \sqrt{F_{\text{Brutus}}^2 + F_{\text{Nitro}}^2 + 2 \cdot F_{\text{Brutus}} \cdot F_{\text{Nitro}} \cdot \cos(\theta)} \]`

Where \(\theta\) is the angle between the forces.

`\[ \theta = 180° - (45° + 30°) \]`

`\[ \theta = 105° \]`

`\[ F_{\text{resultant}} = √(12^2 + 16^2 + 2 \cdot 12 \cdot 16 \cdot \cos(105°)) \]`

`\[ F_{\text{resultant}} \approx 22.61 \ \text{N} \]`

So, the magnitude of the resultant force is approximately
`\(22.61 \ \text{N}\),`and the direction is approximately
`\(105°\)`south of west.

Problem 12:

Given forces as vectors
`\( \mathbf{F}_1 = 9\mathbf{i} - 2\mathbf{j} \), \( \mathbf{F}_2 = -3\mathbf{i} + 10\mathbf{j} \), and \( \mathbf{F}_3 = 18\mathbf{i} - \mathbf{j} \) acting on a \(5 \ \text{kg}\)`object, we can use Newton's second law:

`\[ \mathbf{F}_{\text{net}} = m \mathbf{a} \]`

`\[ \mathbf{a} = \frac{\mathbf{F}_{\text{net}}}{m} \]\[ \mathbf{F}_{\text{net}} = \mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 \]\[ \mathbf{a} = \frac{9\mathbf{i} - 2\mathbf{j} - 3\mathbf{i} + 10\mathbf{j} + 18\mathbf{i} - \mathbf{j}}{5} \]\[ \mathbf{a} = \frac{24\mathbf{i} + 7\mathbf{j}}{5} \]`

The magnitude of acceleration
`(\(|\mathbf{a}|\)`) is given by:

`\[ |\mathbf{a}| = \sqrt{\left((24)/(5)\right)^2 + \left((7)/(5)\right)^2} \]\[ |\mathbf{a}| \approx 5.07 \ \text{m/s}^2 \]`

The direction can be found using:

`\[ \theta = \tan^(-1)\left(\frac{\text{opposite}}{\text{adjacent}}\right) \]\[ \theta = \tan^(-1)\left((7)/(24)\right) \]\[ \theta \approx 16.26° \]`

So, the magnitude of acceleration is approximately
`\(5.07 \ \text{m/s}^2\`), and the direction is approximately
`\(16.26°\)` north of east.

Problem 7:

Given an object of weight
`\(24 \ \text{N}\)` on an inclined plane at
`\(30°\)`, we can find the components of the weight.

`\[ \text{Component down the plane} = W \sin(\theta) \]\[ \text{Component perpendicular to the plane} = W \cos(\theta) \]\[ \text{Component down the plane} = 24 \sin(30°) \]\[ \text{Component down the plane} = 24 \cdot (1)/(2) \]\[ \text{Component down the plane} = 12 \ \text{N} \]\[ \text{Component perpendicular to the plane} = 24 \cos(30°) \]\[ \text{Component perpendicular to the plane} = 24 \cdot (√(3))/(2) \]\[ \text{Component perpendicular to the plane} \approx 20.78 \ \text{N} \]`

So, the component down the plane is \(12 \ \text{N}\), and the component perpendicular to the plane is approximately \(20.78 \ \text{N}\).

Problem 8:

Given vectors
`\( \mathbf{OP} = 20\mathbf{N} \) and \( \mathbf{OQ} = 16\mathbf{N} \)`, we can express these vectors in component form.

a. Expressing
`\( \mathbf{OP} \) and \( \mathbf{OQ} \)` in component form:

`\[ \mathbf{OP} = 20\cos(\theta)\mathbf{i} + 20\sin(\theta)\mathbf{j} \]\[ \mathbf{OQ} = 16\cos(\phi)\mathbf{i} - 16\sin(\phi)\mathbf{j} \]`

b. Calculating the resultant of the two forces:

`\[ \mathbf{OR} = \mathbf{OP} + \mathbf{OQ} \]`

`\[ \mathbf{OR} = (20\cos(\theta) + 16\cos(\phi))\mathbf{i} + (20\sin(\theta) - 16\sin(\phi))\mathbf{j} \]`

The magnitude of the resultant force is given by:

`\[ |\mathbf{OR}| = √((20\cos(\theta) + 16\cos(\phi))^2 + (20\sin(\theta) - 16\sin(\phi))^2) \]`

The direction can be found using:

`\[ \theta_{\text{resultant}} = \tan^(-1)\left((20\sin(\theta) - 16\sin(\phi))/(20\cos(\theta) + 16\cos(\phi))\right) \]`

Make sure to substitute the specific values for
`\( \theta \)` and
`\( \phi \)` into these equations for your specific case.