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PROBLEM 3.151: Solution
PROBLEM 3.151: Solution
PROBLEM 3.151: Solution
151
A 32-lb motor is mounted on the floor. Find the resultant of the weight
and the forces exerted on the belt, and determine where the line of action
of the resultant intersects the floor.
SOLUTION
or R = 185.2 lb 11.84°
Have ΣM O : ΣM O = xRy
∴ − (140 lb ) cos 30° ( 4 + 2 cos 30° ) in. − (140 lb ) sin 30° ( 2 in.) sin 30°
− ( 60 lb )( 2 in.) = x ( 38.0 lb )
1
x= ( −694.97 − 70.0 − 120 ) in.
38.0
Or, resultant intersects the base (x axis) 23.3 in. to the left of the vertical centerline (y axis) of the motor.
PROBLEM 3.152
To loosen a frozen valve, a force F of magnitude 70 lb is applied to the
handle of the valve. Knowing that θ = 25°, M x = −61 lb ⋅ ft, and
M z = −43 lb ⋅ ft, determine θ and d.
SOLUTION
Have ΣM O : rA/O × F = M O
i j k
∴ MO = ( 70 lb ) −4 11 −d in.
−0.90631cos φ −0.42262 0.90631sin φ
634.33
φ = cos −1 = 24.636°
697.86
or φ = 24.6°
1022.90
d = = 34.577 in.
29.583
or d = 34.6 in.
PROBLEM 3.153
When a force F is applied to the handle of the valve shown, its moments
about the x and z axes are, respectively, M x = −77 lb ⋅ ft and
M z = −81 lb ⋅ ft. For d = 27 in., determine the moment M y of F about
the y axis.
SOLUTION
Have ΣM O : rA/O × F = M O
i j k
∴ MO = F −4 11 −27 lb ⋅ in.
cosθ cos φ − sin θ cosθ sin φ
1 Mx
Now, Equation (1) cosθ sin φ = + 27sin θ (4)
11 F
1 Mz
and Equation (3) cosθ cos φ = 4sin θ − (5)
11 F
Substituting Equations (4) and (5) into Equation (2),
1 M 1 M
M y = F −27 4sin θ − z + 4 x + 27sin θ
11 F 11 F
1
or My = ( 27M z + 4M x )
11
PROBLEM 3.153 CONTINUED
27 4
Noting that the ratios and are the ratios of lengths, have
11 11
27 4
My = ( −81 lb ⋅ ft ) + ( −77 lb ⋅ ft ) = 226.82 lb ⋅ ft
11 11
or M y = −227 lb ⋅ ft