Assuming the wedge has an angle of 5°.The minimum value of the force P that is required to begin moving the block up the incline is: 322.84 N.
First step
Using this formula to find the weight of the block
W=mg
W=27×9.81
W=264.87 N
Second step
Angles of friction ∅A and ∅B
∅A=tan^-1(μA)
∅A=tan^-1(0.70)
∅A=34.99°
∅B=tan^-1(μB)
∅B=tan^-1(0.40)
∅B=21.80°
Third step
Equate the sum of forces in m-direction to 0 in order to find the reaction force at B.
∑fm=0
W sin (∅A+20°) + RB cos (∅B+∅A)=0
264.87 sin(34.99°+20°) + RB cos (21.80°+34.99°)=0
216.94+0.5477Rb=0
RB=216.94/0.5477
RB=396.09 N
Fourth step
Equate the sum of forces in x-direction to 0 in order to find force Rc.
∑fx=0
RB cos (∅B) - RC cos (∅B+ 5°)=0
396.09 cos(21.80°) - RC cos (21.80°+5°)=0
RC=396.09 cos(21.80°)/cos(26.80°)
RC=412.02 N
Last step
Equate the sum of forces in y-direction to 0 in order to find force P required to move the block up the incline.
∑fy=0
RB sin (∅B) + RC sin (∅B)-P=0
P=Rb sin (∅B) + RC sin (5°+∅B)
P=396.09 sin(21.80°) +412.02sin (5°+21.80°)
P=322.84 N
Inconclusion the minimum value of the force P that is required to begin moving the block up the incline is: 322.84 N.
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