The following forces act on a hockey puck (round, rubber disk) sitting on a frictionless surface: F_1 = 12 N at 15 degrees; F_2 = 28 N at 125 degrees; and F_3 = 33.3 N at 235 degrees. All the forces are in the plane of the ice. Determine the net force on the puck.
Net force= sum of all.
First, net force at 0 degrees
F@0=12cos15+28cos125+33.3cos235
calculate those, and add.
Then, the forces at 90 deg
F@90=12sin15+28sin125+33.3sin235
calculate, and sum.
Net force= sum of those two forces.
I like to do it this way.
1) angle=arctan F@0/F@90
2) magnitude= F@0/cos(angle)
check that logic.
To determine the net force on the hockey puck, we need to find the vector sum of all the given forces. To do this, we can break each force into its components and then add up the corresponding components.
Let's break down each force into its x-component and y-component using trigonometry.
Force F_1:
Magnitude = 12 N
Angle = 15 degrees
The x-component of F_1, F_1x = F_1*cos(15 degrees)
The y-component of F_1, F_1y = F_1*sin(15 degrees)
Force F_2:
Magnitude = 28 N
Angle = 125 degrees
The x-component of F_2, F_2x = F_2*cos(125 degrees)
The y-component of F_2, F_2y = F_2*sin(125 degrees)
Force F_3:
Magnitude = 33.3 N
Angle = 235 degrees
The x-component of F_3, F_3x = F_3*cos(235 degrees)
The y-component of F_3, F_3y = F_3*sin(235 degrees)
Now, add up the x-components and y-components separately to find the net force:
Net x-component = F_1x + F_2x + F_3x
Net y-component = F_1y + F_2y + F_3y
Finally, determine the magnitude and angle of the net force using the net x-component and net y-component:
Magnitude of net force = sqrt((Net x-component)^2 + (Net y-component)^2)
Angle of net force = arctan(Net y-component / Net x-component)
By following these steps, you can calculate the net force on the hockey puck given the forces acting on it.