Physics

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.
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! :)

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