001 (part 1 of 4)
Call up and to the right positive.
A crate is pulled to the right with a force
of 90.3 N, to the left with a force of 113.3 N,
upward with a force of 589.2 N, and downward
with a force of 238.5 N.
What is the net external force in the x
direction?
Answer in units of N
002 (part 2 of 4)
What is the net external force in the y direction?
Answer in units of N
003 (part 3 of 4)
What is the magnitude of the net external
force on the crate?
Answer in units of N
004 (part 4 of 4)
What is the direction of the net external force
on the crate (as an angle between −180 degrees and
180 degrees, measured from the positive x axis with
counterclockwise positive)?
Answer in units of degrees
Call up and to the right positive.
A crate is pulled to the right with a force
of 90.3 N, to the left with a force of 113.3 N,
upward with a force of 589.2 N, and downward
with a force of 238.5 N.
What is the net external force in the x
direction?
Answer in units of N
002 (part 2 of 4)
What is the net external force in the y direction?
Answer in units of N
003 (part 3 of 4)
What is the magnitude of the net external
force on the crate?
Answer in units of N
004 (part 4 of 4)
What is the direction of the net external force
on the crate (as an angle between −180 degrees and
180 degrees, measured from the positive x axis with
counterclockwise positive)?
Answer in units of degrees
-
Since all four forces are acting at right angles to each other, all we have to do is work out the resultant in the vertical and horizontal directions.
total vertical force = 589.2 - 238.5 = 350.7 N
this acts in the positive y-direction
total horizontal force = - 113.3 + 90.3 = - 23 N
this acts in the negative x-direction.
001. Net force along the x-axis = - 23 N in the negative x-direction
002. Net force along the y-axis = 350.7 N in the positive y-direction
003 From Pythagoras
Net force on crate = √( 350.7^2 + (-23)^2) = 351.45 N
004. From basic trig
Ѳ = arc tan (350.7 / -23) = - 86.25 degrees
This angle is in the second quadrant and is the angle between the resultant and the negative x-axis.
So, going clockwise from the positive x-axis,
the required angle is 180 - 86.25 = 93.75 degrees
total vertical force = 589.2 - 238.5 = 350.7 N
this acts in the positive y-direction
total horizontal force = - 113.3 + 90.3 = - 23 N
this acts in the negative x-direction.
001. Net force along the x-axis = - 23 N in the negative x-direction
002. Net force along the y-axis = 350.7 N in the positive y-direction
003 From Pythagoras
Net force on crate = √( 350.7^2 + (-23)^2) = 351.45 N
004. From basic trig
Ѳ = arc tan (350.7 / -23) = - 86.25 degrees
This angle is in the second quadrant and is the angle between the resultant and the negative x-axis.
So, going clockwise from the positive x-axis,
the required angle is 180 - 86.25 = 93.75 degrees
-
Put everything into vector component form, then add them up
left(-113.3,0)
right(90.3,0)
up(0,589.2)
down(0,238.5)
Final vector components: (-23,350.7)
~You want the net x direction so it would be the x of the final vector component, which is -23 (23 to the left)
~you want the net y direction so it would be the y of the final vector component, which is 350.7 (350.7 up)
~To get the magnitude, you take the pythagoream triangle of the final component
sqrroot((-23)^2+(350.7)^2)=351.5
~to get the direction of the net force, take the tan^-1 (also known as arctan) of the final y component divided by the final x component
tan^-1(350.7/-23)=-86.2 degrees (86.2 degrees under the positive x-axis)
left(-113.3,0)
right(90.3,0)
up(0,589.2)
down(0,238.5)
Final vector components: (-23,350.7)
~You want the net x direction so it would be the x of the final vector component, which is -23 (23 to the left)
~you want the net y direction so it would be the y of the final vector component, which is 350.7 (350.7 up)
~To get the magnitude, you take the pythagoream triangle of the final component
sqrroot((-23)^2+(350.7)^2)=351.5
~to get the direction of the net force, take the tan^-1 (also known as arctan) of the final y component divided by the final x component
tan^-1(350.7/-23)=-86.2 degrees (86.2 degrees under the positive x-axis)
-
net force along y axis is 350.7N. Direction:- positive y-axis
Net force along x-axis is
23N. Direction :- -ve x-axis
So, the resulatant of two forces is (529+122990.49)^2 = 351.45 N
Angle of net external force is theta=tan^-1(23/350.7) + 90 deg = 3.75 + 90 = 93.75 degree from x-axis
Net force along x-axis is
23N. Direction :- -ve x-axis
So, the resulatant of two forces is (529+122990.49)^2 = 351.45 N
Angle of net external force is theta=tan^-1(23/350.7) + 90 deg = 3.75 + 90 = 93.75 degree from x-axis