# An object with a mass of #5 kg# is on a plane with an incline of # - pi/8 #. If it takes #8 N# to start pushing the object down the plane and #2 N# to keep pushing it, what are the coefficients of static and kinetic friction?

I got

To determine the coefficients of (maximum) static and kinetic friction, we can set up a force diagram and take inventory of all of the forces acting on the object.

where

#vecn# is the normal force,#vecF_A# is the applied (pushing) force,#vecf# is the force of friction, and#vecF_G# is the force of gravity, decomposed into its parallel and perpendicular componentsSince we're pushing the object down the plane, I'm going to define down the ramp as the positive direction. This choice is generally up to you.

Therefore, we have:

#color(darkblue)((F_x)_(n et)=sumF_x=F_A+(F_G)_x-f=ma_x)#

#color(darkblue)((F_y)_(n et)=sumF_y=n-(F_G)_y=ma_y)# We are given the following information:

#|->m=5"kg"# #|->theta=-(pi)/8->22.5^o# #|->F_A " to start moving" = 8"N"# #|->F_A " to keep moving" = 2"N"# We'll begin with the coefficient of static friction. The force of static friction is given by:

#color(darkblue)(vecf_(s"max")=mu_svecn)# We are told that it takes

#8"N"# to get the object to start moving, which we interpret as aminimumof#8"N"# . Therefore, there isjust enoughforce applied to move the object, and so there is no acceleration. This means the object is in a state ofdynamic equilibrium.

- This is actually true for both the parallel (x, horizontal) and perpendicular (y, vertical) directions, as the object does not move up and down, meaning that the perpendicular component of gravity is equal and opposite to the normal force.

#color(darkblue)(F_(x" net")=sumF_x=F_A+F_(Gx)-f_(s "max" )=0)#

#color(darkblue)(F_(y" net")=sumF_y=n-F_(Gy)=0)# Hence, we now have:

#F_A+F_(Gx)=f_(s"max")#

#n=F_(Gy)# Using trigonometry, we see that:

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

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

#=>color(blue)((F_G)_x=F_Gsin(theta))# Similarly, we find that

#color(blue)((F_G)_y=F_Gcos(theta))# We also know that

#F_G=mg# , so:

#F_A+mgsin(theta)=mu_(s"max")n#

#=>F_A+mgsin(theta)=mu_(s"max")mgcos(theta)# Solving for

#mu_s# :

#color(blue)(mu_(s"max")=(F_A+mgsin(theta))/(mgcos(theta)))# Substituting in our known values:

#mu_(s"max")=((8"N")+(5"kg")(9.81"m"//"s"^2)sin(22.5^o))/((5"kg")(9.81"m"//"s"^2)cos(22.5^o))#

#=0.591#

#~~color(blue)(0.59)#

- Therefore, the coefficient of maximum static friction is
#~~color(dark blue)(0.59)# , and so the coefficient of static friction is#>=0.59# .For the coefficient of

kineticfriction, we do much of the same, including the assumption of dynamic equilibrium.

#mu_k=(F_A+mgsin(theta))/(mgcos(theta))#

#=((2"N")+(5"kg")(9.81"m"//"s"^2)sin(22.5^o))/((5"kg")(9.81"m"//"s"^2)cos(22.5^o))#

#=0.458#

#~~color(blue)(0.46)#

- Therefore, the coefficient of kinetic friction is
#~~0.46# .We would expect

#mu_s>mu_k# , which works out here, even if slightly.

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The coefficient of static friction is 0.6, and the coefficient of kinetic friction is 0.2.

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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.

- If the length of a #26 cm# spring increases to #46 cm# when a #8 kg# weight is hanging from it, what is the spring's constant?
- If a #1 kg# object moving at #9 m/s# slows down to a halt after moving #40.5 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?
- A block weighing #14 kg# is on a plane with an incline of #pi/6# and friction coefficient of #1/5#. How much force, if any, is necessary to keep the block from sliding down?
- A truck pulls boxes up an incline plane. The truck can exert a maximum force of #2,200 N#. If the plane's incline is #(7 pi )/8 # and the coefficient of friction is #1/3 #, what is the maximum mass that can be pulled up at one time?
- If a #6kg# object moving at #4 m/s# slows down to a halt after moving 12 m, what is the friction coefficient of the surface that the object was moving over?

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