Thread: equilibrium of forces

1. equilibrium of forces

I know i keep posting these sort of questions, its just not clicked yet.

A particle P is on a horizontal plane. Three horizontal forces of magnitude 7N, 10N and 6N, acting in directions as shown in the diagram, are applied to P.

a) show that the resultant of these 3 forces has a magnitude of approximately 6.25N, and find the angle between its direction and that of the 7N force.

b) the particle P has mass 2.5 kg and the coefficient of friction between P and the plane is 0.1. Assuming that the magnitude of the resultant of the three forces is 6.25N, calculate the magnitude of the acceleration of P.

I'm not sure where to even start with this.

I work out

10 = 6cos30

7=6sin30

but what do i do next? i don't really understand the question?

2. Diagram?

3. yeah sorry keeps takin it off

4. Originally Posted by djmccabie
yeah sorry keeps takin it off

start by sketching in the positive x-axis.

$F_x = 6\cos(30) - 10 \approx -4.8 \, N$

$F_y = 7 - 6\sin(30) = 4 \, N$

$|R| = \sqrt{(F_x)^2 + (F_y)^2} \approx 6.25 \, N$

direction of the resultant is in quad II ...

$\theta = \arctan\left(\frac{F_y}{F_x}\right) + 180^{\circ} \approx 140^{\circ}$ ... which is $50^{\circ}$ relative to the 7 N force.

5. is that always how you find the resultant force by using pythagoras?

When you find the angle how come you +180??

thanks for your help

6. is that always how you find the resultant force by using pythagoras?

yes, the magnitude of the resultant vector is the vector sum of its two perpendicular components

When you find the angle how come you +180??

because the inverse tangent function of a negative value yields a negative angle in quad IV ... since the resultant angle is in quad II, it requires the 180 degree addition.

7. so say it moved in a direction in the third quadrant you add 270?

do you always inverse tan ?

8. Originally Posted by djmccabie
is that always how you find the resultant force by using pythagoras?

When you find the angle how come you +180??

thanks for your help
I don't know if you are allowed to use vectors. If so: Place the origin at the particle P. Then you have the forces:

$\overrightarrow{F_1} = (-10,0)$

$\overrightarrow{F_2} = (0,7)$

$\overrightarrow{F_3}=(6\cdot \cos(330^\circ), 6 \cdot \sin(330^\circ)) =( 3\sqrt{3}, -3)$

Angles are measured between the positive x-axis and the vector anti-clockwise. Thus you have to add 180° to the angle of 150°.

The resulting force is the sum of all three vectors:

$\overrightarrow{F_R} = (-10,0)+ (0,7)+ ( 3\sqrt{3}, -3) = (3\sqrt{3}-10, 4)$

The magnitude of the force correspond with the length of the resulting vector:

$|\overrightarrow{F_R}| = \sqrt{(3\sqrt{3}-10)^2 + 4^2} = \sqrt{143-60\sqrt{3}} \approx 6.25116...$

To calculate the angle between two vectors $\vec a$ and $\vec b$ use the formula:

$\cos(\vec a, \vec b) = \dfrac{\vec a \cdot \vec b}{|\vec a| \cdot |\vec b|}$

With your question you have:

$\cos(\alpha) = \dfrac{(0,7) \cdot (3\sqrt{3}-10, 4)}{7 \cdot \sqrt{143-60\sqrt{3}}} \approx 0.639882~\implies~\alpha \approx 50.217^\circ$

9. I like your reply it is very detailed and understandable. Is the formula
$

\cos(\vec a, \vec b) = \dfrac{\vec a \cdot \vec b}{|\vec a| \cdot |\vec b|}
$

true for all possible values? Is it something i should memorise or are there any other solutions such as the inverse tan? is it the same as working out a line on an argand diagram ie |Z| and argZ ?

thanks for your help

10. Originally Posted by djmccabie
I like your reply it is very detailed and understandable. Is the formula
$

\cos(\vec a, \vec b) = \dfrac{\vec a \cdot \vec b}{|\vec a| \cdot |\vec b|}
$

true for all possible values? Yes
Is it something i should memorise or are there any other solutions such as the inverse tan? is it the same as working out a line on an argand diagram ie |Z| and argZ ?

thanks for your help
This formula $

\cos(\vec a, \vec b) = \dfrac{\vec a \cdot \vec b}{|\vec a| \cdot |\vec b|}
$
is nothing but the definition of the dot-product of two vectors:

$\vec a \cdot \vec b = |\vec a| \cdot |\vec b| \cdot \cos(\vec a, \vec b)$

Now solve this equation for $\cos(\vec a, \vec b)$ and you'll get the formula I used to calculate the angle.