11.70 One end of a uniform Figure P11.70 meter stick is placed against a vertica

Question

11.70 One end of a uniform Figure P11.70 meter stick is placed against a vertical wall (Fig. P11.70). The other end is held by a light weight cord that makes an angle with the stick. The coefficient of static friction between the end of the meter stick and the wall is 0.40 (a) What is the maximum value the angle can have if the stick is to remain in equilibrium? (b) Let the angle be 15°. 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? (c) When = 15°, how large must the coefficient of static friction be so that the block can be attached 10 cm from the left end of the stick with out causing it to slip

Explanation / Answer

T is the tension in the cord and th is the angle


The normal force at the wall is T*cos(th)

sum torque at the center
mu is the friction coefficient

T*cos(th)*mu/2=T*sin(th)/2
tan(th)=mu
or th=atan(0.4)
th=21.8 degrees

B)

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 --- 1

Summation of vertical forces = 0

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

Summation of horizontal forces = 0

Fx = T(cos theta)

the frictional force = (mu)(Normal force)
Fy = (mu)Fx

or

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

Fx = Fy/mu = T(cos theta)

Fy = mu(T)(cos theta)

Substitute "Fy = mu(T)(cos theta)" in Equation 2

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

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

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

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

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 15/[(0.4 * cos 15) + sin 15]} - 0.5

x = 0.3023 meter = 30.23 cm

The block should be 0.3023m (or 30.23 cm) from the wall for the system to be in equilibrium.



C)

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

Solving for "mu" from the above,

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


Again, substituting approriate values (theta = 15 degrees and x = 0.10 m),

mu = tan 15[(2/(0.10 + 0.5)) - 1]

mu = 0.6252

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