,max
N
s
F
f
=
1
μ
s
=
tan
θ
.
Therefore,
μ
s
= 0.35.
70. (a) The angle between the beam and the floor is
sin
−
1
(
d
/
L
)= sin
−
1
(1.5/2.5) = 37
°
,
so that the angle between the beam and the weight vector
W
→
of the beam is 53
°
.
With
L
=
2.5 m being the length of beam, and choosing the axis of rotation to be at the base,
Σ
τ
z
=
0
¡
PL
–
W
©
¨
§
¹
¸
·
L
2
sin 53
°
=
0
Thus,
P
= ½
W
sin 53
°
= 200 N.
(b) Note that
P
→
+
W
→
= (200
∠
90
°
) + (500
∠
–127
°
) = (360
∠
–146
°
)
using magnitudeangle notation (with angles measured relative to the beam, where
"uphill" along the beam would correspond to 0
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

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