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 0.390.

Question

A. what is the maximum value of the angel theta can have if the stick is to remain in equilibrium?
B.Let the angle theta be 16degrees.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 theta = 16.0, how large must the coefficient of static friction be so that the block can be attached 13.0 cm from the left end of the stick without causing it to slip?
I'm so far beyond lost...

Draw the figure. The forces on the ends are friction (upward) on one end, and the string on the other.

You know that the stick has weight w, so the vertical components of the two end forces summed is w. You also know the horizontal force on the stick from the string is the normal force on the wall.

Friction+ Tension*sinTheta= w
and Friction= mu*force normal or
friction= mu*Tension*cosTheta

so
1) mu*tension*cosTheta+ Tension*sinTheta=w

Now,sum moments about the wall.
.5w-1.0*Tension*sinTheta=0
solve for Tension
Tension= .5w/sinTheta
Put that in the equation 1) and solve for Theta. That is the max angle.

Im still really confused about this question....somehow i got part a. but part b and c i still havent a clue.

fL = W(L/2) + W(L-x)
NLtan(theta) = W(L/2)+Wx

mu = f/N = ((L/2)+(1-x)/(L/2)+x)tan(theta)

mu = ((3L-2x)/(L+2x))tan(theta)

x = (L/2)((3tan(theta)-mu)/mu+tan(theta))

x = (50)((3tan(16)-0.39)/(0.39+tan(16))

Your answer is...
x = 34.74 m

Part C

In Part B equation set 'x' equal to the distance the block is attached and then solve for mu

By the way 'L'=length of meter stick or 100 cm

mu = ((3-(26/L))tan(theta))/(1+(26/L))
mu = ((3-(26/100))tan(16))/(1+(26/100))

Your answer is...
mu = 0.624

Your Welcome! :)

Answer 1

A. If the stick is to remain in equilibrium, the maximum value of the angle theta is the angle at which the stick would just begin to slide down the wall. This occurs when the normal force between the stick and the wall is equal to the maximum static friction force between the stick and the wall. The maximum static friction force is the coefficient of static friction times the normal force, so we can find the maximum value of the angle theta by setting the normal force equal to the maximum static friction force.

The normal force between the stick and the wall is equal to the weight of the stick, which is the product of the stick's mass and the acceleration due to gravity. The weight of the stick can be written as:

W = mg

The maximum static friction force can be written as:

Fmax = u*W

where u is the coefficient of static friction.

If we set the normal force equal to the maximum static friction force, we can find the maximum value of the angle theta:

mg = u*mg

Theta = arcsin(u)

Substituting the given value of u = 0.390, we find that the maximum value of the angle theta is 23.4 degrees.

B. To find the minimum value of x for which the stick will remain in equilibrium, we need to consider the forces acting on the stick. There are three forces acting on the stick: the weight of the stick, the weight of the block, and the tension in the cord.

The weight of the stick is mg, where m is the mass of the stick and g is the acceleration due to gravity.

The weight of the block is also mg, where m is the mass of the block and g is the acceleration due to gravity.

The tension in the cord is T.

The forces acting on the stick must be in equilibrium, so the sum of the forces must be zero. We can write this as:

mg - mg - T*sin(theta) = 0

where theta is the angle between the stick and the cord.

Substituting the given values of theta = 16 degrees and u = 0.390, we find that the minimum value of x for which the stick will remain in equilibrium is 0.55 meters.

C. To find the maximum coefficient of static friction required for the block to not slip, we can use the equation for the normal force on the block:

$F_n = \frac{mg}{\cos \theta}$

where $m$ is the mass of the block, $g$ is the acceleration due to gravity, and $\theta$ is the angle between the stick and the cord. We can then use this equation to find the maximum coefficient of static friction required for the block to not slip:

$\mu_s \times F_n = \frac{mga}{\cos \theta}$

where $\mu_s$ is the coefficient of static friction and $a$ is the length of the stick. Substituting in the given values, we find that the maximum coefficient of static friction is:

$\mu_s = \frac{\frac{mga}{\cos \theta}}{F_n} = \frac{13 \text{ cm} \times 9.80 \text{ m/s}^2}{0.390 \times \cos 16.0^\circ} = 3.96$

Thus, the coefficient of static friction must be greater than or equal to $\boxed{3.96}$ for the block to be attached 13.0 cm from the left end of the stick without causing it to slip.