Statics Wedges b1j1n

WHAT ARE WEDGES?

W

A wedge is a simple machine in which a small force P is used to lift a large weight W. To determine the force required to push the wedge in or out, it is necessary to draw FBDs of the wedge and the object on top of it. It is easier to start with a FBD of the wedge since you know the direction of its motion. Note that: a) the friction forces are always in the direction opposite to the motion, or impending motion, of the wedge; b) the friction forces are along the ing surfaces; and, c) the normal forces are perpendicular to the ing surfaces.

WHAT ARE WEDGES? (CONTINUATION)

Next, a FBD of the object on top of the wedge is drawn. Please note that: a) at the ing surfaces between the wedge and the object the forces are equal in magnitude and opposite in direction to those on the wedge; and, b) all other forces acting on the object should be shown.

WHAT ARE WEDGES? (CONTINUATION) W

If the object is to be lowered, then the wedge needs to be pulled out. If the value of the force P needed to remove the wedge is positive, then the wedge is selflocking, i.e., it will not come out on its own. However, if the value of P is negative, or zero, then the wedge will come out on its own unless a force is applied to keep the wedge in place. This can happen if the coefficient of friction is small or the wedge angle  is large.

WEDGES?

• Wedges - simple machines used to raise heavy loads. • Force required to lift block is significantly less than block weight. • Friction prevents wedge from sliding out. • Want to find minimum force P to raise block.

• Block as free-body

å Fx = 0 :

- N1 + m s N 2 = 0

å Fy = 0 : - W - m s N1 + N 2 = 0

• Wedge as free-body

å Fx = 0 :

- m s N 2 - N3 (m s cos 6° - sin 6°) +P=0

å Fy = 0 :

- N 2 + N3 (cos 6° - m s sin 6°) = 0

EXAMPLE  Determine the force P required to start the wedge. The angle of friction for all surfaces in is 15°.

SOLUTION  From the FBD of the block to the right ΣFV=0 R1cos15∘=R2sin15∘+200 R1=0.2679R2+207.06 ΣFH=0ΣFH=0 R2cos15∘=R1sin15∘+50 R2cos15∘=(0.2679R2+207. 06)sin15∘+50 0.8966R2=103.59 R2=115.54 kN

SOLUTION  From the FBD of the wedge to the left ΣFH=0 R3cos30∘=R2cos15∘ R3cos30∘=115.54cos15∘ R3=128.87 kN ΣFV=0 P=R2sin15∘+R3sin30∘ P=115.54sin15∘+128.87si n30∘ P=94.34 kN answer

EXAMPLE  The wedge is used to level the floor of a building. For the floor loading shown, determine the horizontal force P that must be applied to move the wedge forward. The coef ficient of static friction between the wedge and the two surfaces of is μs. Neglect the size and weight of the wedge and the thickness of the beam. Given: F1 = 2 kN F2 = 4 kN F3 = 4 kN F4 = 2 kN μ s = 0.25

a=3m b=2m c=3m θ = 15 deg

SOLUTION

 (F1 − N1)(a + b + c) + F2(b + c) + F3 c = 0

 N2 cos(θ) − μ sN2 sin(θ) − N1 = 0  μ sN1 + μ sN2 cos(θ) + N2 sin(θ ) − P = 0

N1 = 6000 N N2 = 6660 N P = 4830 N

Ans. Ans. Ans.

EXERCISE FOR WEDGES  Determine the minimum weight of block B that will keep it at rest while a force P starts blocks A up the incline surface of B. The weight of A is 100 lb and the angle of friction for all surfaces in is 15 °.

EXERCISE FOR WEDGES  From the FBD of block A ΣFV=0 R1cos 45∘=100 R1=141 .42 lb  From the FBD of block B ΣFH=0 R2sin15∘=R1sin45∘ R2sin15∘=141 .42sin45∘ R2=386.37 lb ΣFV=0 WB+R1cos45∘=R2cos15∘ WB+141 .42cos45∘=386.37cos15∘ WB=273.20 lb answer