Asked by Amy
                A block of mass 3 kg is at rest on an incline. The coefficient of state friction between the block and incline is 0.34. What is the maximum angle of the incline such that the block will not slide down the plane?
 
 
            
            
        Answers
                    Answered by
            Henry
            
    Wb = m * g = 3kg * 9.8N/kg = 29.4 N. = Wt. of the block.
Fb = 29.4N. @ Ao. = Force of the block.
Fp = 29.4*sinA = Force parallel to the incline.
Fv = 29.4*cosA = Force perpendicular to the incline.
Fs = u*Fv = 0.34 * 29.4*cosA = 10*cosA.
Fp-Fs = m*a.
29.4*sinA-10*cosA = m*0 = 0
29.4*sinA = 10*cosA
Divide both sides by cosA:
29.4*(sinA/cosA) = 10
Replace sinA/cosA with tanA:
29.4*tanA = 10
tanA = 0.34
A = 18.8o
    
Fb = 29.4N. @ Ao. = Force of the block.
Fp = 29.4*sinA = Force parallel to the incline.
Fv = 29.4*cosA = Force perpendicular to the incline.
Fs = u*Fv = 0.34 * 29.4*cosA = 10*cosA.
Fp-Fs = m*a.
29.4*sinA-10*cosA = m*0 = 0
29.4*sinA = 10*cosA
Divide both sides by cosA:
29.4*(sinA/cosA) = 10
Replace sinA/cosA with tanA:
29.4*tanA = 10
tanA = 0.34
A = 18.8o
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