# Vectors help!!!! (1 Viewer)

#### Run hard@thehsc

##### Well-Known Member

How do you solve this question? (at least the first part for question 19...)

#### ExtremelyBoredUser

##### Bored Uni Student
View attachment 34166
View attachment 34167
How do you solve this question? (at least the first part for question 19...)

the red vectors are the sin components of mg, which is the weight force of the objects.

a.

$\bg_white F = T - mg\sin(\theta)$
$\bg_white ma = T - mg\sin(\theta)$
$\bg_white ma + mg\sin(\theta) = T$
$\bg_white 3a + 3g\sin(\theta) = T_{1}$ as m = 3

b.

$\bg_white F = mg\sin(2\theta) - T$ double the angle as in the diagram
$\bg_white T = mg\sin(2\theta) + F$
$\bg_white T = mg\sin(2\theta) + ma$
$\bg_white T_{2} = 2g\sin(2\theta) + 2a$ m=2 on this section

c.

a = 0 when the system is in equilibrium and tension in the rope is the same
$\bg_white T_{2} = T_{1}$
$\bg_white 2g\sin(2\theta) + 2a = 3a + 3g\sin(\theta)$
$\bg_white 2g\sin(2\theta)= 3g\sin(\theta)$
$\bg_white 2\sin(2\theta)= 3\sin(\theta)$ divide by g
$\bg_white 2\sin(2\theta) - 3\sin(\theta) = 0$
$\bg_white 4\sin(\theta)\times\cos(\theta) - 3\sin(\theta) = 0$ double angle formula
$\bg_white \sin(\theta)\times(4\cos(\theta) - 3) = 0$

$\bg_white 4\cos(\theta) = 3$
$\bg_white cos(\theta) = \frac{3}{4}$

as required

Last edited:

#### vishnay

##### God
View attachment 34171
the red vectors are the sin components of mg, which is the weight force of the objects.

a.

$\bg_white F = T - mgsin(\theta)$
$\bg_white ma = T - mgsin(\theta)$
$\bg_white ma + mgsin(\theta) = T$
$\bg_white 3a + 3gsin(\theta) = T_{1}$ as m = 3

b.

$\bg_white F = mgsin(2\theta) - T$ double the angle as in the diagram
$\bg_white T = mgsin(2\theta) + F$
$\bg_white T = mgsin(2\theta) + ma$
$\bg_white T_{2} = 2gsin(2\theta) + 2a$ m=2 on this section

c.

a = 0 when the system is in equilibrium and tension in the rope is the same
$\bg_white T_{2} = T_{1}$
$\bg_white 2gsin(2\theta) + 2a = 3a + 3gsin(\theta)$
$\bg_white 2gsin(2\theta)= 3gsin(\theta)$
$\bg_white 2sin(2\theta)= 3sin(\theta)$ divide by g
$\bg_white 2sin(2\theta) - 3sin(\theta) = 0$
$\bg_white 4sin(\theta)*cos(\theta) - 3sin(\theta) = 0$ double angle formula
$\bg_white sin(\theta) * (4cos(\theta) - 3) = 0$

$\bg_white 4cos(\theta) = 3$
$\bg_white cos(\theta) = \frac{3}{4}$

as required
for trig functions u can do \sin like this $\bg_white \sin \text{or} \cos$

and also u dont have to do asterisk just do \times like this $\bg_white \times$

#### ExtremelyBoredUser

##### Bored Uni Student
for trig functions u can do \sin like this $\bg_white \sin \text{or} \cos$

and also u dont have to do asterisk just do \times like this $\bg_white \times$
thx

#### Run hard@thehsc

##### Well-Known Member
@ExtremelyBoredUser Thanks for your help. I got the first two after trying and realised the third one after seeing your working out!