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
Morgan
Aug 11, 2017

#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)#

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