Determine the force P required to move the block A of 5000 N weight up the inclined plane, coefficient of friction between all contact surfaces is 0.25. Neglect the weight of the wedge and the wedge angle is 15 degrees.

Given : Weight of block A = 5000 N

μ_{s}=0.25

Wedge angle = 15^{º}

To find : Force P required to move block A up the inclined plane

#### Solution

The impending motion of block A is to move up

The block A is in equilibrium

N_{1,}N_{2},N_{3} are the normal reactions

Fs_{1} = μ1N_{1} = 0.25N_{1}F_{s2} = μ_{2}N_{2} = 0.25N_{2}Fs_{3} = μ_{3}N_{3} = 0.25N_{3}Applying the conditions of equilibrium

ΣFy = 0

∴ -5000 + N_{1} cos 60 – Fs_{1} sin 60 – Fs_{2} sin 15 + N_{2} cos 15 = 0

∴ N_{1} x 0.5 – 0.25N_{1} x 0.866 -0.25 N_{2} x 0.2588 + N_{2} x 0.9659 = 5000 (From 1)

∴ 0.2835 N_{1} +0.9012 N_{2} = 5000 ……………..(2)

Applying the conditions of equilibrium

ΣFx = 0

∴N_{1} sin 60 +Fs_{1} cos 60 –Fs_{2} cos 15 – N_{2} sin 15 =0

∴ 0.866 N_{1} + 0.25 x N_{1} x 0.5 -0.25 x N_{2} x 0.9659 –N_{2} x 0.2588 = 0(From 1)

∴ 0.991 N_{1} -0.5003 N_{2} = 0

Solving equation, no 2 and 3

N_{1} = 2417.0851 N

N_{2} = 4787.79 N

The impending motion of block B is towards left Block B is in equilibrium. Applying the conditions of equilibrium

ΣFy = 0

∴ N_{3} + F_{s2} sin 15 –N_{2} cos 15 = 0

∴N_{3} + 0.25N_{2} x 0.2588 – N_{2} x 0.9659 = 0

∴ N_{3} – 0.9012 N_{2} = 0

∴ N_{3}= 0.9012 x 4787.79 = 4314.7563**Applying conditions of equilibrium**ΣFx = 0

∴ -P + F

_{s3}+F

_{s2}cos 15 +N

_{2}sin 5 =0

∴ 0.25 N

_{3}+0.25 N

_{2}x 0.9659 +N

_{2}x 0.2588 =P

∴ P = 0.25 N

_{3}+0.5003 N

_{2}= 0.25 X 4314.7563 + 0.5003 x 4787.79 = 3474 N

**The force P required to move the block A of weight 5000 N up the inclined plane is P=3474 N**