Are These Grade 12 Vector Solutions Correct?

[madeeeeee]

Plz check if my answers to these questions are correct. Thank you.

1. Express CA + TC – RC + RT as a single vector.

-> CA+TC-CR+RT
-> CA+TR+RT

Answer: CA+TT

2. A 150.0 kg mass is suspended from a ceiling by two cords that make angles of 32 degreeand 40 degree with the ceiling. Find the tension in each of the cords.

Answer: For T1 i got 120.8 kg and for T2 i got 133.7 kg

3.Find a single vector equivalent to each of the following expressions.

a) 2(-2,3,3) + 3(-1,3,-1)

Answer: (-4, 6, 6) + (-3, 9, -3)
: (-7, 15, 3)


b) 2(i-2j+3k) - 3(-j+4j-3k)

Answer: 3i-4j+6k+3j-12j+9k
: (3i-16j+15k)

 

[Mark44]

Plz check if my answers to these questions are correct. Thank you.

1. Express CA + TC – RC + RT as a single vector.

What do CA, TC, and so on represent? Are these the displacement vectors between two points?

-> CA+TC-CR+RT
-> CA+TR+RT

How did you go from - RC to -CR? If the first of these represents the vector from point R to point C, then the vector CR points the opposite direction.

Answer: CA+TT

What would TT represent?

2. A 150.0 kg mass is suspended from a ceiling by two cords that make angles of 32 degreeand 40 degree with the ceiling. Find the tension in each of the cords.

Answer: For T1 i got 120.8 kg and for T2 i got 133.7 kg

These are pretty close to what I got (120.8 and 133.75, which would round to 133.8). The tensions should be in Newtons, though, not kg.

3.Find a single vector equivalent to each of the following expressions.

a) 2(-2,3,3) + 3(-1,3,-1)

Answer: (-4, 6, 6) + (-3, 9, -3)
: (-7, 15, 3)

This is correct.

b) 2(i-2j+3k) - 3(-j+4j-3k)

Do you have a typo in the 2nd vector? It looks like it should be <-i + 4j -3k>

Answer: 3i-4j+6k+3j-12j+9k
: (3i-16j+15k)

 

[madeeeeee]

1. Yes they are displacement vectors, and i learned that if there is a negative vector than change it to positive by switching the letters and once we have everything positive we can solve.

3. For 3 b i did make a typo :) it is supposed to be 2(i-2j+3k) - 3(-i+4j-3k), sorry

 

[madeeeeee]

i have another question that i haven't been able to solve, actually i don't know how to approach it:

Question: A log is dragged 3.5 m along the horizontal ground, then up a 4.5 m ramp inclined at 25 degrees to the ground. The pulling force is consistently 240.0 N at an angle of 45 degrees up from horizontal. Calculate the total work done in moving this log.

 

[Mark44]

1. Yes they are displacement vectors, and i learned that if there is a negative vector than change it to positive by switching the letters and once we have everything positive we can solve.



3. For 3 b i did make a typo :) it is supposed to be 2(i-2j+3k) - 3(-i+4j-3k), sorry

In the first problem, where you have

Answer: CA+TT

TT would be the vector from T to T. I.e., this is the zero vector.

 

[Mark44]

i have another question that i haven't been able to solve, actually i don't know how to approach it:

Question: A log is dragged 3.5 m along the horizontal ground, then up a 4.5 m ramp inclined at 25 degrees to the ground. The pulling force is consistently 240.0 N at an angle of 45 degrees up from horizontal. Calculate the total work done in moving this log.

Break this into two parts: the work necessary to move the log the first 3.5 m, and the work necessary to pull the log up the 4.5 m ramp. For each part, the work is the component of the force in the direction of motion, times the distance.

 

[madeeeeee]

240*3.5*cos(45°)+240*4.5*cos(20°) joules
=1609 J

 

[madeeeeee]

An airplane heads N80W with an airspeed of 680.0 km/h. Measurements made from the ground indicate that the plane’s groundspeed is 650.0 km/h at N85W. Find the windspeed and wind direction. [7 marks]

So a bearing of N80°W = 90+80 =170°
and a bearing of N85°W = 90+85 = 175°

Let a = vector indicating airplane's speed and direction.
Let w = vector indicating wind's speed and direction.
Let g = vector indicating plane's speed and direction from the ground

a + w = g
w = g - a
w = 650 (cos(175), sin(175)) - 680 (cos(170), sin(170))
w = (650cos(175)-680cos(170), 650sin(175)-680sin(170))
w = (22.142718, -61.429528)

|w| = √(22.142718^2 + 61.429528^2) = 65.298445

w = 65.298445 (cos(θ), sin(θ))

sin(θ) = -61.429528/65.298445 = -0.94075
cos(θ) = 22.142718/65.298445 = 0.3391
tan(θ) = -61.429528/22.142718

Since cos(θ) > 0 and sin(θ) < 0, then terminal point of θ is in quadrant IV
arctan has range between -90 and 90 (i.e. values in quadrant IV and quadrant I)
So we can just take arctan to find θ
(Otherwise, when θ is in QII or QIII, then we have to add 180 to arctan)

θ = arctan (-61.429528/22.142718) = -70.18

Changing this back into a bearing, we get S19.82°ESo windspeed = 65.3 km/h at S19.82°EIs this right and is there a simpler way to approach this