An object with a mass of # 15 kg# is on a ramp at an incline of #pi/8 #. If the object is being pushed up the ramp with a force of # 7 N#, what is the minimum coefficient of static friction needed for the object to remain put?
We are given the following information:
#>m=15"kg"# #>theta=pi/8# #>F_a=7"N"#
Two important points about static friction...

Recall that it is the force of maximum static friction which is proportional to the magnitude of the normal force,
#n# , and that#f_s<=f_(s"max")# 
The direction of the force of static friction acting on an object is opposite the direction that the object would move if the force of friction was not present
We can calculate the coefficient of friction at the maximum force of static friction, where our forces are in equilibrium. Keep in mind that an object remains at rest as long as
The force of friction in question then should be less than
Conclusion:
A diagram of the situation:
Let's define up the incline as positive.
Summing the parallel (x) and perpendicular (y) components of our forces, we have:
#(1)" "color(darkblue)(sumF_x=f_s+F_a(F_G)_x=cancel(ma_x)=0)#
#(2)" "color(darkblue)(sumF_y=n(F_G)_y=cancel(ma_y)=0)#
We also know that:
#(3)" "color(darkblue)(f_(s"max")=mu_sn)#
#(4)" "color(darkblue)(F_G=mg)#
Substituting equations
#(5)" "color(darkblue)(mu_s(F_G)_y+F_a(F_G)_x=0)#
We can rearrange equation
#=>(6)" "color(darkblue)(mu_s=((F_G)_xF_a)/(F_G)_y)#
We can find the components of the force of gravity using trigonometry.
#sin(theta)=(F_G)_x/F_G#
#=>(F_G)_x=F_Gsin(theta)#
Similarly,
By equation

#(F_G)_x=mgsin(theta)# 
#(F_G)_y=mgcos(theta)# We can substitute these equations into equation
#(6)# :#color(crimson)(mu_(s"max")=(mgsin(theta)F_a)/(mgcos(theta))# Now we can substitute in our known values:
#=>mu_(s"max")=((15"kg")(9.81"m"//"s"^2)sin(pi/8)7"N")/((15"kg")(9.81"m"//"s"^2)cos(pi/8))# #=0.363# #:.# The coefficient of static friction is#<=0.363# .
By signing up, you agree to our Terms of Service and Privacy Policy
The minimum coefficient of static friction needed for the object to remain put is approximately 0.268.
By signing up, you agree to our Terms of Service and Privacy Policy
When evaluating a onesided 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 onesided 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 onesided 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 onesided 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.
 A bullet of mass 15g leaves the barrel of a rifle with a speed of 800 m/s. If the length of the barrel is 75 cm, what is the the average force that accelerates the bullet?
 An object with a mass of #2 kg# is on a surface with a kinetic friction coefficient of # 7 #. How much force is necessary to accelerate the object horizontally at #6 m/s^2#?
 An object with a mass of #6 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 1+3cotx #. How much work would it take to move the object over #x in [(pi)/6, (3pi)/8], where x is in meters?
 An object with a mass of #14 kg# is on a plane with an incline of #  pi/3 #. If it takes #12 N# to start pushing the object down the plane and #11 N# to keep pushing it, what are the coefficients of static and kinetic friction?
 An object with a mass of #5 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= xe^x+x #. How much work would it take to move the object over #x in [1, 2], where x is in meters?
 98% accuracy study help
 Covers math, physics, chemistry, biology, and more
 Stepbystep, indepth guides
 Readily available 24/7