Question

11.70 ... One end of a uniform
meter stick is placed against a
vertical wall (Fig. P11.70). 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 0.40.
(a) What is the maximum value
the angle can have if the stick
is to remain in equilibrium? (b) Let the angle be 15

Answer 1

let w = weight of meter stick = weight of block
T = tension in cord
at equilibrium, CCW moment = CW moments
(Tsin 15)(1.0) = w(0.5) + w(x)
0.276T = 0.5w + w(x)

FF = force of friction on meter stick at wall
FF = (0.40)(Normal force)
Normal force = T cos 16 = 0.961T
FF = (0.40)(0.961T) = 0.384T

sum of forces upward = sum of forces downward
FF + T sin 16 = w + w
0.384T + 0.276T = 2w
0.66T = 2w
w = 0.33T

sustitute for w in moment equality above:
0.276T = 0.5 w + w(x)
0.276T = (0.5)(0.33T) + (0.33T)(x)
0.276T - 0.165T = 0.33T(x)
cancel T's
0.276 - 0.165 = 0.33x
0.111 = 0.33x
x = 0.111/0.33 = 0.34 m ANS-1

IF x = 0.14 m
at equilibrium, CCW moment = CW moments
(Tsin 16)(1.0) = w(0.5) + w(0.14)
0.276T = 0.5w + w(0.14)
0.276T = 0.64w
w = 0.276T/0.64 = 0.431T

FF = 2w - T sin 16
FF = 2(0.431)T - 0.276T
FF = 0.862T - 0.276T
FF = 0.586T =