One end of a uniform meter stick is placed against a vertical wall. The other en

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 ? with the stick. The coefficient of static friction between the end of the meter stick and the wall is 0.356.

a) What is the maximum value the angle theta can have if the stick is to remain in equilibrium?

b) Let the angle between the cord and the stick is theta = 14.4?. 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 = 14.4?, how large must the coefficient of static friction be so that the block can be attached 13.8cm from the left end of the stick without causing it to slip?

Explanation / Answer

mu=tan(theta)

theta=tan-1(0.356)

=19.595degrees

B)

theta=14degrees

let m = mass of meter stick and mass of block
let T = tension in cord
summing moments about the contact point on the wall with CW positive
the lever arm distance for the cord tension is the perpendicular line from the cord through the contact point or (1)sin18

9.81m(0.5) + 9.81m(x) - Tsin14 = 0
9.81m(0.5 + x) - Tsin14 = 0
9.81m(0.5 + x) = Tsin14
in terms of x
T = 9.81m(0.5 + x) / sin14

summing forces in the horizontal direction Where Fr = normal force of stick on wall
Fr - Tcos14 = 0
Fr = Tcos14
in terms of x
Fr = (9.81m(0.5 + x) / sin14)cos14

summing forces in the vertical direction
0.356Fr + Tsin14- 2(9.81)m = 0
in terms of x
0.356(9.81m(0.5 + x) / sin14)cos14 + (9.81m(0.5 + x) / sin14) sin14 - 2(9.81)m = 0

as each term has a 9.81m in the numerator we can divide that out leaving

0.356((0.5 + x) / sin14)cos14 + ((0.5 + x) / sin14) sin14- 2 = 0

multiply each term by sin18 to get rid of denominators

0.356(0.5 + x)cos14 + (0.5 + x)sin14 - 2sin14 = 0
(0.5 + x)(0.356cos14 + sin14) = 2sin14
(0.5 + x) = 2sin14 / (0.39cos14 + sin14)
x = 2sin14 / (0.356cos14 + sin14) - 0.5
x = 0.323 meters

From the moment equation, as x increases, T increases. It then follows that Fr and thus the frictional force also increase so this is the minimum x to maintain static equilibrium.


Part b)

we can use the equation already developed to find a new k coefficient of friction if we plug in x = 0.138 meters.
(0.5 + 0.138)(kcos14.4 + sin14.4) = 2sin14.4
kcos14.4 + sin14.4 = (2sin14.4 / 0.638)
kcos14.4 = (2sin14.4/ 0.638) - sin14.4
k = ((2sin14.4 / 0.638) - sin14.4) / cos14.4
k = 0.548

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