The wedge is being clamped in place by the wood jaws the jaw

The wedge is being clamped in place by the wood jaws. (the jaw on the left weighs 10 lb, the jaw on the right 20 lb. the concrete wedge weighs 40 lb) Find beta which moves the wedge up and down. Discuss how changing the _2 values on either side would change this problem

Solution

Assume wedge moves up:

Let the incline have normal N1, friction towards the incline downwards f1
Let the vertical have normal N2, friction downwards = f2

Forcec Balance parallel and perpendicular the incline

f1 + f2cos(beta) - N2sin(beta) = 0
N1 = N2 cos(beta) + f2sin(beta)
f2 = N2mu2
f1 = N1mu1
N1mu1 - N2sin(beta) + N2mu2cos(beta) = 0
N1 = N2cos(beta) + N2mu2sin(beta) = N2(cos(beta) + mu2Sin(beta))
N2(cos(beta) + mu2sin(beta))mu1 = N2(sin(beta) + mu2cos(beta))
beta = arctan([mu1 - mu2]/[1-mu1*mu2])
now, mu1 = tan(15) = 0.267, mu2 = tan(12) = 0.2125
beta = 3.30 degrees

Assume wedge moves down:

Let the incline have normal N1, friction towards the incline upwards f1
Let the vertical have normal N2, friction upwards = f2

Forcec Balance parallel and perpendicular the incline

-f1 - f2cos(beta) - N2sin(beta) = 0
N1 = N2 cos(beta) - f2sin(beta)
f2 = N2mu2
f1 = N1mu1
-N1mu1 - N2sin(beta) - N2mu2cos(beta) = 0
N1 = -N2cos(beta) - N2mu2sin(beta) = -N2(cos(beta) + mu2Sin(beta))
N2(cos(beta) + mu2sin(beta))mu1 = N2(sin(beta) + mu2cos(beta))
beta = arctan([mu1 - mu2]/[1-mu1*mu2])
now, mu1 = tan(15) = 0.267, mu2 = tan(12) = 0.2125
beta = 3.30 degrees

Now tan(beta) = [mu1 - mu2]/[1-mu1*mu2]
sec^2(beta)d(beta) = [[1-mu1*mu2] - [mu1 - mu2][-m2]]dmu1/[1-mu1*mu2]^2
sec^2(beta)d(beta) = [[1-tan(15)tan(12)] - [tan(15) - tan(12)][-tan(12)]]dmu1/[1-tan(12)tan(15)]^2 = 1.06*dmu1
sec^2(beta)d(beta) = [-[1-mu1*mu2] - [mu1 - mu2][-m1]]dmu2/[1-mu1*mu2]^2
sec^2(beta)d(beta) = [-[1-tan(15)tan(12)] - [tan(15) - tan(12)][-tan(15)]]dmu2/[1-tan(12)tan(15)]^2 = [-0.943 + 0.01484]dmu2/(0.889) = -1.044dmu2

hence, beta increases with increase in mu1, but decreases with increase in mu2