Forces in Equilibrium?

Question

Forces in Equilibrium?

Three horizontal forces of magnitudes $(F ) N$ , $63N$ and $25N$ act at $O$ , the origin of the x-axis and y-axis.
The forces are in equilibrium. The force of magnitude $(F) N$ makes an angle $theta$ anticlockwise with the positive $x$ -axis. The force of magnitude $63N$ acts along the negative $y$ -axis. The force of magnitude $25N$ acts at $tan^-1(0.75)$ clockwise from the negative $x$ -axis. Find the value of $F$ and the value of $tantheta$ .

1 Answer

Impact of this question

3719 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-11T03:19:36

$F_1=52"N, " tan(theta)=2.40$

Explanation:

We can use trigonometry, vector decomposition, and Newton's second law to find the desired values.

We have the following information:

$|->F_2=63"N"$
$|->F_3=25"N"$
$|->theta_1=arctan(0.75)$

We want to find $theta_2$ and $F_1$ .

Here is a diagram:

enter image source here

The forces are in a state of dynamic equilibrium, meaning that the acceleration is zero. Therefore, we can sum the parallel and perpendicular components of the three forces as follows:

$(1)color(darkblue)(sumF_x=F_(1x)-F_(3x)=ma_x=0)$

$(2)color(darkblue)(sumF_y=F_(1y)+F_(3y)-F_2=ma_y=0)$

Let's look at $F_3$ first.

Using trigonometry, we can find the parallel (x, horizontal) and perpendicular (y, vertical) components of $F_3$ .

$sin(theta)="opposite"/"hypotenuse"$

$=>sin(theta_1)=(F_(3y))/F_3$

$=>F_(3y)=F_3sin(theta_1)$

$=>=25sin(arctan(0.75))$

$=>=15$

$:.color(darkblue)(F_(3y)=15"N")$

Similarly, we can find $F_(3x)$ :

$F_(3x)=F_3cos(theta_1)$

$=25cos(arctan(0.75))$

$=20$

$:.color(darkblue)(F_(3x)=20"N")$

Then, using equation $(1)$ we can derive an equation for $F_(1x)$ :

$F_(1x)=F_(3x)$

$=>F_1cos(theta_2)=20$

$color(darkblue)(F_1=20/cos(theta_2))$

Using equation $(2)$ , we can derive an equation for $F_(1y)$ :

$F_(1y)=F_2-F_(3y)$

$=>F_1sin(theta_2)=63-15$

$=>color(darkblue)(F_1=48/sin(theta_2))$

We now have two equations for $F_1$ , which we can set equal to solve for $theta_2$ .

$20/cos(theta_2)=48/sin(theta_2)$

$=>sin(theta_2)/cos(theta_2)=48/20$

$=>tan(theta_2)=12/5$

$=>theta_2=arctan(12/5)$

$=>color(darkblue)(theta_2=67.38^o)$

We can put this value back into either of our equations for $F_1$ to find its magnitude:

$F_1=20/cos(theta_2)$

$=>=20/(cos(67.38^o))$

$=>=52$

$:.color(crimson)(F_1=52"N")$

Finally, we can find $tan(theta_2)$ :

$=>tan(67.38^o)=color(crimson)(2.40)$

Science

Math

Humanities

... and beyond

Forces in Equilibrium?

1 Answer

Explanation:

Related questions

Impact of this question