Question #2212e

2 Answers
Nathan L.
Aug 19, 2017

#"magnitude" = 2390# #"N"#

#"direction" = 82.4^"o"# measured anticlockwise from the positive #x#-direction

Explanation:

We're asked to find the resultant (net) force acting on the halfback.

To do this, let's first split each force into its respective components:

#x_"Jerome" = (1230color(white)(l)"N")cos(53^"o") = ul(740color(white)(l)"N"#

#y_"Jerome" = (1230color(white)(l)"N")sin(53^"o") = ul(982color(white)(l)"N"#

#x_"Michael" = (1450color(white)(l)"N")cos(107^"o") = ul(-424color(white)(l)"N"#

#y_"Michael" = (1450color(white)(l)"N")sin(107^"o") = ul(1387color(white)(l)"N"#

To find the components of the net force, we add the respective components:

#x_"net" = 740color(white)(l)"N" - 424color(white)(l)"N" = color(red)(ul(316color(white)(l)"N"#

#y_"net" = 982color(white)(l)"N" + 1387color(white)(l)"N" = color(green)(ul(2369color(white)(l)"N"#

The magnitude of the resultant force is given by

#color(blue)(F_"net") = sqrt((x_"net")^2 + (y_"net")^2) = sqrt((color(red)(316color(white)(l)"N"))^2 + (color(green)(2369color(white)(l)"N"))^2) = color(blue)(ulbar(|stackrel(" ")(" "2390color(white)(l)"N"" ")|)#

And the direction is

#color(blue)(theta) = arctan((y_"net")/(x_"net")) = arctan((color(green)(2369color(white)(l)"N"))/(color(red)(316color(white)(l)"N"))) = color(blue)(ulbar(|stackrel(" ")(" "82.4^"o"" ")|)#

measured anticlockwise from the positive #x#-direction.

Morgan
Aug 19, 2017

#F_"net"=2390"N"# at #82^o# North of East.

Explanation:

Force Diagram:

enter image source here

  • Note that the angles are taken counterclockwise from the east, which is equivalent to taking them as counterclockwise from the positive x-axis, as is typical.

We are given the following information:

  • #|->F_J=1230"N"#
  • #|->theta_J=53^o#
  • #|->F_M=1450"N"#
  • #|->theta_M=107^o#

We can use our force diagram to generate statements for the net force parallel (x, horizontal) and perpendicular (y, vertical).

#=>sumF_x=F_"Jx"-F_"Mx"#
#=>sumF_y=F_"Jy"+F_"My"#

  • Note that I have taken East/right and North/upwards to be positive directions, which is how the signs of each force component are determined in the above net force statements.

We are given the magnitudes of each force as well as the angle at which they occur, so we can use that information along with some basic trigonometry to calculate the components of #F_J# and #F_M#

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

From the diagram above, we see:

#sin(theta_J)=F_"Jy"/F_J#

#=>color(blue)(F_"Jy"=F_Jsin(theta_J))#

Similarly:

  • #color(blue)(F_"Jx"=F_Jcos(theta_J))#.
  • #color(blue)(F_"Mx"=F_Mcos(theta_M))#.
  • #color(blue)(F_"My"=F_Msin(theta_M))#.

Returning to our net force statements, we now have:

#color(blue)(F_(x" net")=F_Jcos(theta_J)-F_Mcos(theta_M))#

#color(blue)(F_(y" net")=F_Jsin(theta_J)+F_Msin(theta_M))#

We can now substitute in our known values to calculate the net components:

#F_(x" net")=1230"N"cos(53^o)-1450"N"cos(73^o)#

#=>color(blue)(=316.294"N")#

#F_(y" net")=1230"N"sin(53^o)+1450"N"sin(73^o)#

#=>color(blue)(=2368.964"N")#

  • Note that I am using #theta_M=180-107=73^o#. This is the equivalent angle which forms an imaginary right triangle with West/the negative y-axis.

The magnitude of the net force is then given by:

#F_"net"=sqrt(F_x^2+F_y^2)#

#=>=sqrt((316.294"N")^2+(2368.964"N")^2)#

#=>color(blue)(=2390"N")#

If you want to know the direction as well as the magnitude:

#tan(theta)="opposite"/"adjacent"#

In this case, we have:

#tan(theta_"net")=(F_y/F_x)#

#=>theta_"net"=arctan(F_y/F_x)#

#=>=arctan(2368.964/316.294)#

#=>=82.395^o#

#=>color(blue)(~~82^o)#

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