# 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?

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

#(F_G)# into its parallel and perpendicular components. This can be done using basic trigonometry, noting that 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

#F_(Gy)=mgsin(theta)# .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

#m# .

#=>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")#

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To find the maximum mass that can be pulled up the incline, we need to consider the forces acting on the system. The force of gravity pulling the box down the incline is countered by the force exerted by the truck, along with the force of friction opposing the motion.

The force of gravity ( F_{\text{gravity}} ) can be calculated as:

[ F_{\text{gravity}} = mg ]

The force exerted by the truck ( F_{\text{truck}} ) can be calculated as:

[ F_{\text{truck}} = F_{\text{max}} = 3500 , \text{N} ]

The force of friction ( F_{\text{friction}} ) can be calculated using the equation:

[ F_{\text{friction}} = \mu \cdot F_{\text{normal}} ]

Where ( \mu ) is the coefficient of friction and ( F_{\text{normal}} ) is the normal force. Since the incline plane is inclined at an angle ( \theta ), the normal force is equal to:

[ F_{\text{normal}} = mg \cdot \cos(\theta) ]

So, the force of friction becomes:

[ F_{\text{friction}} = \mu \cdot mg \cdot \cos(\theta) ]

To find the maximum mass that can be pulled up, we need to determine the mass when the net force up the incline equals zero:

[ F_{\text{truck}} - F_{\text{friction}} = 0 ]

Substitute the expressions for ( F_{\text{truck}} ) and ( F_{\text{friction}} ) into the equation:

[ 3500 , \text{N} - \mu \cdot mg \cdot \cos(\theta) = 0 ]

[ 3500 , \text{N} - \frac{9}{4} \cdot mg \cdot \cos\left(\frac{3\pi}{8}\right) = 0 ]

Now, solve for ( m ), the mass:

[ m = \frac{3500}{\frac{9}{4} \cdot \cos\left(\frac{3\pi}{8}\right)} ]

[ m = \frac{3500 \cdot 4}{9 \cdot \cos\left(\frac{3\pi}{8}\right)} ]

[ m \approx \frac{14000}{9 \cdot \cos\left(\frac{3\pi}{8}\right)} ]

[ m \approx \frac{14000}{9 \cdot \cos(67.5^\circ)} ]

[ m \approx \frac{14000}{9 \cdot 0.2588} ]

[ m \approx \frac{14000}{2.3292} ]

[ m \approx 6004.8 , \text{kg} ]

So, the maximum mass that can be pulled up at one time is approximately 6004.8 kilograms.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- An object with a mass of # 15 kg# is lying on a surface and is compressing a horizontal spring by #10 cm#. If the spring's constant is # 8 (kg)/s^2#, what is the minimum value of the surface's coefficient of static friction?
- If the length of a #17 cm# spring increases to #63 cm# when a #3 kg# weight is hanging from it, what is the spring's constant?
- If an object is moving at #3 m/s# over a surface with a kinetic friction coefficient of #u_k=5 /g#, how far will the object continue to move?
- If an object is moving at #150 m/s# over a surface with a kinetic friction coefficient of #u_k=15 /g#, how far will the object continue to move?
- An object, previously at rest, slides #4 m# down a ramp, with an incline of #pi/12 #, and then slides horizontally on the floor for another #3 m#. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

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