A level Maths part 05

• May 23rd 2007, 06:50 AM
A level Maths part 05
thanks for everyone helping me ;) i am really greatful ,and the explations are makiing understand more thank you ...i need help in this homework too ,and i need answers for all please..there is another part continued.

regards
• May 24th 2007, 05:04 AM
CaptainBlack
Quote:

thanks for everyone helping me ;) i am really greatful ,and the explations are makiing understand more thank you ...i need help in this homework too ,and i need answers for all please..there is another part continued.

regards

http://www.mathhelpforum.com/math-he...untitled-1.jpg

1. Vector sum of forces:

$
\bold{V} = (2\bold{i} + \bold{j}) + (3\bold{i} - 2\bold{j}) + (-\bold{i} -3\bold{j}) = (2+3-1) \bold{i} +(1-2-3)\bold{j} = 4 \bold{i} -4\bold{j} \mbox{ N}
$

The the acceleration $\bold{a} = \bold{V}/m$

2. let the tension be $\bold{T}$ then the total force on the first particel, and hence the acceleration is:

$
\bold{T}-m_1\bold{g} = m_1 \bold{a_1}
$

and for the second particle:

$
\bold{T}-m_2\bold{g} = m_2 \bold{a_2}
$

but $\bold{a_2} = -\bold{a_1}$ so:

$
\bold{T}/m_2-\bold{g} = -[\bold{T}/m_1-\bold{g}]
$

which can then b solved for $\bold{T}$

RonL
• May 24th 2007, 05:22 AM
topsquark
Start by drawing a Free-Body-Diagram for the ball.

I have a +x axis to the right and a +y axis upward. There is a weight (w) acting straight downward, a tension (T) acting at 50 degrees above the -x axis, and an applied force (F) of 50 N acting in the +x direction.

Since the ball is in equilibrium we know that
$\sum F_x = 0$
$\sum F_y = 0$

So:
$\sum F_x = -Tcos(50) + F = 0$
which implies
$T = \frac{F}{cos(50)} = \frac{50 \, N}{cos(50)} = 77.7862 \, N$

and
$\sum F_y = Tsin(50) - w = 0$
which implies
$w = Tsin(50) = \frac{F}{cos(50)} \cdot sin(50) = F tan(50)$

$= (50 \, N)tan(50) = 59.5877 \, N$

-Dan
• May 24th 2007, 01:41 PM