Determine the largest angle - wedge

[blackandyello]

Homework Statement

A small wedge is placed beneath corner B of the 4000-kg block of marble.
Determine the largest angle θ for which the wedge is self-locking; that is, the
wedge will not slide out from under the block. Neglect the mass of the wedge and
the small angle between surface AB and the horizontal.

http://screensnapr.com/e/527Zk6.jpg

Homework Equations

using the FBD of the whole composite body.

summation of forces on x = 0
0.2Na - 0.2Nc = 0

summation of forces on y = 0
-39240 + Na + Nc = 0

solving, it yields = Na = 19620 Nc = 19620

The Attempt at a Solution

now, we take the fbd of just the wedge

summation of forces along x = 0
Nb cos(90-θ) -fb cos(θ) - fc = 0

summation of forces along y = 0
-Nb sin(90-θ) - fb sin(θ) + Nc = 0

dividing those 2 eqns yields:

-tan(90-θ) + tan(θ) - (Nc/fc)
then since we have Nc and Fc (using f = uN) where u = 0.2

and θ yields arounds 70 degrees which is incorrect (the book says 18.15 degrees)

thanks

 

[anathema]

for the help, but I still don't understand why the correct answer is 18.15 degrees. Can you explain it further?

Sure. Let's start by looking at the free body diagram of just the wedge:



There are three forces acting on the wedge: the normal force from the block (N), the frictional force from the block (f), and the normal force from the ground (Nc). We can break down these forces into components along the x and y directions:



Note that we have chosen the x-axis to be parallel to the surface of the wedge, and the y-axis to be perpendicular to the surface of the wedge. We can then set up the following equations:

ΣFx = 0: Ncos(90-θ) - fcos(θ) = 0
ΣFy = 0: Nsin(90-θ) + fsin(θ) - Nc = 0

We also know that the force of friction is given by f = μN, where μ is the coefficient of friction and N is the normal force. In this case, μ = 0.2 and N = Nc. So we can substitute this into our equations:

ΣFx = 0: Ncos(90-θ) - 0.2Ncos(θ) = 0
ΣFy = 0: Nsin(90-θ) + 0.2Nsin(θ) - Nc = 0

We can then solve for N and Nc in terms of θ:

N = 0.2Ncos(θ)/cos(90-θ)
Nc = Nsin(90-θ) + 0.2Nsin(θ)

Now, we know that the wedge is self-locking when it does not slide out from under the block, which means that the frictional force must be equal to or greater than the force of the block pushing down the wedge (N). So we can set up the following inequality:

f ≥ N

Substituting in our equations for N and Nc:

0.2Ncos(θ) ≥ N
0.2Ncos(θ) ≥ 0.2Ncos(θ)/cos(90-θ)

We can simplify this to:

cos(