Question

1) (10 points total) A system securing a large weather balloon is shown with the balloon's upward force acting at attachment ring, A (balloon not shown). Point "O" is at the origin of the xyz coordinate system. Note the orientation of the coordinate system. B -2 ft -3 ft F 1 ft 6 ft 1 ft y 3 ft The vector expressions for the position vectors for cables "AC" and "AD" are given by $\vec{r}_{AC} = -3\hat{i} - \hat{j} - 6\hat{k}$ ft and $\vec{r}_{AD} = 2\hat{i} + 3\hat{j} - 6\hat{k}$ ft respectively. Ez a) (2 points) What is the length of cable "AD"? $|AD| = \sqrt{-2^2 + 3^2 + (-6)^2} = 7$ Ez b) (2 points) What is the unit vector in the direction of cable "AD"? $\vec{u} = \frac{\vec{AD}}{|AD|} = \frac{1}{7}(-2\hat{i} + 3\hat{j} - 6\hat{k}) = \frac{-2}{7}\hat{i} + \frac{3}{7}\hat{j} - \frac{6}{7}\hat{k}$ c) (2 points) What is the angle between cable "AD" and the z-axis? $\vec{A} = x$ $\vec{B} = y$ $cos\gamma$ d) (4 points) What is the angle between cables "AD" and "AC"?

          1) (10 points total) A system securing a large weather balloon is shown with the
balloon's upward force acting at attachment ring, A (balloon not shown).
Point "O" is at the origin of the xyz coordinate system. Note the orientation of
the coordinate system.
B
-2 ft
-3 ft
F
1 ft
6 ft
1 ft
y
3 ft
The vector expressions for the position vectors for cables "AC" and "AD" are
given by $\vec{r}_{AC} = -3\hat{i} - \hat{j} - 6\hat{k}$ ft and $\vec{r}_{AD} = 2\hat{i} + 3\hat{j} - 6\hat{k}$ ft respectively.
Ez a) (2 points) What is the length of cable "AD"?
$|AD| = \sqrt{-2^2 + 3^2 + (-6)^2} = 7$
Ez b) (2 points) What is the unit vector in the direction of cable "AD"?
$\vec{u} = \frac{\vec{AD}}{|AD|} = \frac{1}{7}(-2\hat{i} + 3\hat{j} - 6\hat{k}) = \frac{-2}{7}\hat{i} + \frac{3}{7}\hat{j} - \frac{6}{7}\hat{k}$
c) (2 points) What is the angle between cable "AD" and the z-axis?
$\vec{A} = x$
$\vec{B} = y$
$cos\gamma$
d) (4 points) What is the angle between cables "AD" and "AC"?

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1) (10 points total) A system securing a large weather balloon is shown with the
balloon's upward force acting at attachment ring, A (balloon not shown).
Point "O" is at the origin of the xyz coordinate system. Note the orientation of
the coordinate system.
B
-2 ft
-3 ft
F
1 ft
6 ft
1 ft
y
3 ft
The vector expressions for the position vectors for cables "AC" and "AD" are
given by r⃗AC = -3î - ĵ - 6k̂ ft and r⃗AD = 2î + 3ĵ - 6k̂ ft respectively.
Ez a) (2 points) What is the length of cable "AD"?
|AD| = √(-2^2 + 3^2 + (-6)^2) = 7
Ez b) (2 points) What is the unit vector in the direction of cable "AD"?
u⃗ = (A⃗D⃗)/(|AD|) = (1)/(7)(-2î + 3ĵ - 6k̂) = (-2)/(7)î + (3)/(7)ĵ - (6)/(7)k̂
c) (2 points) What is the angle between cable "AD" and the z-axis?
A⃗ = x
B⃗ = y
cosγ
d) (4 points) What is the angle between cables "AD" and "AC"?

Added by Susan M.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Madhur L

06/15/2023

Step 1

To find the length of cable AD, we need to calculate the magnitude of the vector AD. The magnitude of a vector is given by the formula ||v|| = sqrt(v1^2 + v2^2 + v3^2 + ... + vn^2), where v1, v2, v3, ..., vn are the components of the vector. Given that AD = V-2 + Show more…

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Texts: The Coordinate System The vector expressions for the position vectors for cables AC and AD are: AC = a2 AD = V-2 + 3 - 6 Question 1: What is the length of cable AD? 1AD = V-2 + 3 - 6 Question 2: What is the unit vector in the direction of cable AD? AD U = ADI Question 3: What is the angle between cables AD and AC? P = y22 dson d

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Transcript

00:01 Hello, in the question we have given three cables a, b and c are attached to a hook.

00:07 So these are the three vectors a, b and c.

00:11 So r is the resultant which is given in the question.

00:14 So we have to determine the magnitude of this vector b and this angle phi.

00:20 So now in order to do this we will use the, we will see the horizontal components.

00:30 So we will have horizontal components of vectors.

00:42 So now if i drop a perpendicular over here, so of a, so i will see, note that this a will be pointing in the negative x direction.

00:54 So it will be, and this magnitude will be a cos of 20.

00:59 So this, if i just drop over here, if i make this triangle, so now this is the right angle triangle which i am considering.

01:06 Now this side i can use the trigonometric identities and i will note that this will be minus of a cos of 20 degrees.

01:18 Then i will have this c.

01:21 So now c also, if i drop this perpendicular over here like this, so this will be sin phi.

01:30 So c sin phi.

01:32 This is the right angle triangle which i am considering and this vectors i can translate.

01:37 So i can translate it over here and i can see that the c is going in the x direction.

01:42 So this will be c sin of phi that is equal to minus r.

01:50 So in the same way, if i take the projection of this r along this, so i will get this r sin of 23.

02:02 Now let us plug the value.

02:04 A, we know what it is.

02:05 A is 65 cos of 20 plus c, we know it is 40 sin of phi that is equal to minus r is 100 sin of 23.

02:23 So if i calculate this, so i will get sin of phi.

02:29 So that is equal to minus of 100 sin of 23 is minus of 39 .07 and then this i will bring over here.

02:40 So it will become plus 65 cos 20 degree is 61 .08 divided by this 40.

02:50 So i will get sin phi as equal to 0 .55025.

03:01 So phi will be equal to sin inverse of this 0 .55025.

03:13 So now if we calculate, so this angle phi, we will get it as 33 .4 degrees.

03:22 So this is the first part...

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