# A truck pulls boxes up an incline plane. The truck can exert a maximum force of #3,500 N#. If the plane's incline is #(3 pi )/8 # and the coefficient of friction is #9/4 #, what is the maximum mass that can be pulled up at one time?

##### 1 Answer

#### Answer:

#### Explanation:

We can use **Newton's second law** to derive an equation for the maximum mass that the truck can pull up the incline at a time.

(I make the assumption that the truck and mass are always simultaneously on the incline and that whatever mechanism the used to connect the truck to the box is irrelevant.)

**Force diagram for a box:**

- Where
#vecn# is the normal force,#vecf_k# is the force of kinetic friction,#vecT# is the force of the truck, and#F_G# is the force of gravity, decomposed into its parallel (x, horizontal) and perpendicular (y, vertical) components.

**We have the following information:**

#|->mu_k=9/4# #|->theta=(3pi)/8# #|->F_T=3500"N"#

**Relevant equations:**

#f_k=mu_kn# #F_G=mg#

We can set up statements of the net force using the above information and diagram. Due to the nature of the situation, we will assume *dynamic equilibrium,* where the truck pulls the box at a constant velocity.

#sumF_x=F_T-f_k-F_(Gx)=0#

#sumF_y=n-F_(Gy)=0#

We will have to decompose the gravitational force **the angle between the force of gravity vector and the vertical is equal to the angle of incline.**

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

#=>sin(theta)=F_(Gx)/F_G#

#=>F_(Gx)=F_Gsin(theta)#

#=>=mgsin(theta)#

Similarly, we can use the cosine function to show that

We now have:

#n=mgcos(theta)#

Therefore:

#color(darkblue)(F_T-mu_kmgcos(theta)-mgsin(theta)=0)#

We can rearrange the above equation to solve for

#=>color(darkblue)(m=F_T/(g(sintheta+mu_kcostheta)))#

Substituting in our known values:

#m=(3500"N")/((9.81"m"//"s"^2)(sin((3pi)/8)+9/4cos((3pi)/8))#

#=>m=199.885"kg"#

#=>color(darkblue)(m~~200"kg")#