Answer:
a) Fb= 275.77 lb Fc= 142.75 lb
b) M = -779.97 lb.ft (i.e. 779.97 lb.ft in clockwise direction)
c) Fax = 195 lb
Fay = 337.75 lb
Fbx = 195 lb
Fby = 195 lb
Explanation:
Question: Three tugboats are used to turn a barge in a narrow channel. To avoid producing any net translation of the barge, the forces applied should be couples. The tugboat at point A applies a 390 lb force.
(a) Determine FB and FC so that only couples are applied.
(b) Using your answers to Part (a), determine the resultant couple moment that is produced.
(c) Resolve the forces at A and B into x and y components, and identify the pairs of forces that constitute couples.
Solution:
<u>For this problem Right hand side is positive X direction and Upwards is positive Y direction. Couples and moments will be considered positive in counterclockwise direction.</u>
<u />
a) For no translation condition
∑
& ∑
Hence,


and

Inserting the value of
and solving the remaining equations simultaneously yields (magnitudes),
b) Summing up moments

(i.e. 779.97 lb.ft clockwise)
c)



Answer:
f=1.59 Hz
Explanation:
Given that
Simple undamped system means ,system does not consists any damper.If system consists damper then it is damped spring mass system.
Velocity = 100 mm/s
Maximum amplitude = 10 mm
We know that for a simple undamped system spring mass system

now by putting the values

100 = ω x 10
ω = 10 rad/s
We also know that
ω=2π f
10 = 2 x π x f
f=1.59 Hz
So the natural frequency will be f=1.59 Hz.
Answer:
i) σ1 = 133.5 MPa
σ2 = -2427 MPa
ii) 78.89 MPa
Explanation:
Given data:
ε1 = 0.0020 and ε2 = –0.0010
E = 71 GPa
v = 0.35
<u>i) Determine the biaxial stresses σ1 and σ2 using the relations below</u>
ε1 = σ1 / E - v (σ2 / E) -----( 1 )
ε2 = σ2 / E - v (σ1 / E) -------( 2 )
resolving equations 1 and 2
σ1 = E / 1 - v^2 { ε1 + vε2 } ---- ( 3 )
σ2 = E / 1 - v^2 { ε2 + vε1 } ----- ( 4 )
input the given data into equation 3 and equation 4
σ1 = 133.5 MPa
σ2 = -2427 MPa
<u>ii) Calculate the value of the maximum shear stress ( Zmax )</u>
Zmax = ( σ1 - σ2 ) / 2
= 133.5 - ( - 2427 ) / 2
= 78.89 MPa
Answer:
A vehicle is considered to be legally parked if it is parked 20 feet (6 m) or more from a pedestrian crosswalk or a marked or unmarked intersection.
Explanation:
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