Equilibrium problem need help?

Question

One end of a uniform meter stick is placed against a vertical wall. The other end is held by a lightweight cord that makes an angle theta with the stick. The coefficient of static friction between the end of the meter stick and the wall is 0.39

1) Let the angle between the cord and the stick is theta = 16 degrees. A block of the same weight as the meter stick is suspended from the stick, as shown, at a distance x from the wall. What is the minimum value of x for which the stick will remain in equilibrium?

2) When theta = 16 degrees, how large must the coefficient of static friction be so that the block can be attached 13 cm from the left end of the stick without causing it to slip?

Answer 1

Let

Fx = horizontal force exerted by the wall on the stick

Fy = frictional force (vertical force) exerted by wall on the stick

T = tension in the lightweight cord (assume to be massless)

m = mass of the stick = mass of the block

g = acceleration due to gravity

x = distance where the block is to be suspended along the stick for the system to be in equilibrium

mu = coefficient of friction = 0.39 (given)

Summation of moments about the point where the meter stick touches the wall = 0 (for system to be in equilibrium)

T(sin theta) = mgx + 0.5mg  --- Call this Equation 1

Summation of vertical forces = 0

Fy + T(sin theta) = 2mg --- Call this Equation 2

Summation of horizontal forces = 0

Fx = T(cos theta)

NOTE that, by definition, the frictional force = (mu)(Normal force) and in this problem,

Fy = (mu)Fx

or

Fx = Fy/mu and since Fx = T(cos theta)  -- from Equation 3 --

Fx = Fy/mu = T(cos theta)

and solving for Fy,

Fy = mu(T)(cos theta)

Substitute  "Fy = mu(T)(cos theta)" in Equation 2 and you will have

mu(T)(cos theta) + T(sin theta) = 2mg

and solving for "T",

T = 2mg/[mu(cos theta) + sin theta]

Now, substitute "T = 2mg/[mu(cos theta) + sin theta]" in Equation 1 and you will have

(2mg*sin theta)/[mu(cos theta) + sin theta] = mgx + 0.5mg  

NOTE that the factors "mg" appear on both sides of the equation, hence will simply cancel out. Rearranging the above equation to solve for "x",

x = {(2*sin theta)/[mu(cos theta) + sin theta]} - 0.5

Substituting appropriate values,

x = {2 * sin 16/[(0.39 * cos 16) + sin 16]} - 0.5

x = 0.847 - 0.5

x = 0.347 meter = 34.7 cm

The block should be 0.347 m (or 34.7 cm) from the wall for the system to be in equilibrium.

**************************************...

This derived relationship (from Problem 1)

x = {(2*sin theta)/[mu(cos theta) + sin theta]} - 0.5

will have to be modified and solved for "mu" for Problem 2.

Solving for "mu" from the above,

mu = tan theta[(2/(x + 0.5)) - 1]

NOTE -- I trust that, on your own, you can arrive at the above derived formula for "mu" (as this is basic algebraic manipulation).

Again, substituting approriate values (theta = 16 degrees and x = 0.13 m),

mu = tan 16[(2/(0.13 + 0.5)) - 1]

mu = 0.62