# An object with a mass of #5 kg# is on a plane with an incline of #pi/8 #. If the object is being pushed up the plane with # 8 N # of force, what is the net force on the object?

##### 1 Answer

*down* the plane)

#### Explanation:

**We are given the following information:**

#|->m=5"kg"# #|->theta=pi/8# #|->F_p=8"N"#

**A diagram of the situation:**

Where

#vecn# is the normal force,#vecF_G# is the force of gravity, decomposed into its parallel and perpendicular components, and#vecF_p# is the pushing force.

Note:I will assume a "frictionless" incline based on the information provided.

Let's define **up the incline as positive**.

We can calculate the **net force** on the object by summing the parallel (x, horizontal) and perpendicular (y, vertical) components of our forces. We have:

#F_(x" net")=sumF_x=F_p-(F_G)_x=ma_x#

#F_(y" net")=sumF_y=n-(F_G)_y=ma_y#

We can assume *dynamic equilibrium* vertically, where

#color(darkblue)(F_(x" net")=F_p-(F_G)_x)#

#color(darkblue)(F_(y" net")=0#

Since we're given **angle between the force of gravity and the vertical is equal to the angle of incline.**

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

#=>sin(theta)=(F_G)_x/F_G#

#=>(F_G)_x=F_Gsin(theta)#

Because

#=>color(darkblue)((F_G)_x=mgsin(theta))#

Which gives us:

#color(darkblue)(F_(x" net")=F_p-mgsin(theta))#

#color(darkblue)(F_(y" net")=0#

We can now calculate the parallel and perpendicular components of the net force.

#F_(x" net")=F_p-mgsin(theta)#

#=8"N"-(5"kg")(9.81"m"//"s"^2)sin(pi/8)#

#=-10.77"N"#

#~~-11"N"#

- Note that the negative sign indicates
*direction*. We defined*up the ramp*as positive, so a negative parallel component of the net force tells us that the net force horizontally is*down the ramp.*