A crate is pulled to the right with a force of 80 N, to the left with a force of 125.8 N, upward with a force of 615.4 N, and downward with a force of 248 N. What is the net external force in the x
direction? y direction??
Also....
What is the magnitude of the net external force on the crate?
What is the direction of the net external force on the crate (as an angle between −180◦ and 180◦, measured from the positive x axis with counterclockwise positive)
You have to decide which is + x, and +y.
Lets use to the right as +x, and up as +y
x direction: 80-125N
y direction: 615.4-248N
angle : arc tan yforce/x force
A body of weight 500N is lying on a rough plane inclined at angle 28 with the horizontal. Determine the minimum and maximum value of P,for which the equiliarium can exist if the angle of friction is 25
To find the net external force in the x-direction, we need to sum all the forces acting in the x-direction. The positive x-direction is to the right and the negative x-direction is to the left.
Given:
- Force pulling to the right: 80 N (positive x-direction)
- Force pulling to the left: 125.8 N (negative x-direction)
To find the net external force in the x-direction, we subtract the force pulling to the left from the force pulling to the right:
Net external force in the x-direction = Force pulling to the right - Force pulling to the left
= 80 N - 125.8 N
= -45.8 N (negative sign indicates the force is acting to the left)
Therefore, the net external force in the x-direction is -45.8 N.
Similarly, to find the net external force in the y-direction, we sum all the forces acting in the y-direction. The positive y-direction is upward, and the negative y-direction is downward.
Given:
- Force pulling upward: 615.4 N (positive y-direction)
- Force pulling downward: 248 N (negative y-direction)
Net external force in the y-direction = Force pulling upward - Force pulling downward
= 615.4 N - 248 N
= 367.4 N (positive value indicates the force is acting upward)
Therefore, the net external force in the y-direction is 367.4 N.
To calculate the magnitude of the net external force on the crate, you can use the Pythagorean theorem:
Magnitude of net external force = sqrt((net external force in the x-direction)^2 + (net external force in the y-direction)^2)
= sqrt((-45.8 N)^2 + (367.4 N)^2)
= sqrt(2102.64 N^2 + 134884.76 N^2)
= sqrt(137987.4 N^2)
= 371.6 N (approximately)
Therefore, the magnitude of the net external force on the crate is approximately 371.6 N.
To find the direction of the net external force on the crate, you can use trigonometry. The angle is measured counterclockwise from the positive x-axis.
Direction of the net external force = arctan(net external force in the y-direction / net external force in the x-direction)
= arctan(367.4 N / -45.8 N)
= arctan(-8.016)
The arctan result is -82.12°. Since the angle is measured counterclockwise, the direction of the net external force on the crate is approximately -82.12°.