The horizontal beam in (Figure 1) weighs 190 N. and its center of grav

eozoischgc

eozoischgc

Answered question

2022-01-16

The horizontal beam in (Figure 1) weighs 190 N. and its center of gravity is at its center.
a) Find the tension in the cable.
b) Find the horizontal component of the force exerted on the beam at the wall.
c) Find the vertical component of the force exerted on the beam at the wall.
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Answer & Explanation

levurdondishav4

levurdondishav4

Beginner2022-01-17Added 38 answers

Step 1
Apply equailibrium condition of torques about the pivot point by taking the counter clockwise torques as positive. Thus,
Tsinθ(L)(300N)(L)(190N)(L2)=0
T=190N(L2)+300N(L)Lsinθ
The horizontal componenet and vertical component of force are,
Fx=Tcosθ=(685N)(45)=527N
Fy=(300+190)Tsinθ=300+190(685N)(35)=95N
sukljama2

sukljama2

Beginner2022-01-18Added 32 answers

Step 1
Take moments about the hinge on the wall
(2×190)+(4×300)(4×TsinA)=0
T - tension
A - angle between beam and cable
sinA=35
380+1200(125)T=0
1580=(125)T
7900=12T
T=658.33N
Step 2
The only other horizontal force on the beam is from the cable.
In equilibrium
force at wall = Hor. force in cable
Horizontal force=Horizontal component of T=TcosA
cosA=45
Horizontal force =658.33×45
=526.66N
Step 3
In equilibrium:
sum of vertical forces = 0
300+190TsinAY=0
490(658.33×35)Y=0
490394.88=Y
Y=95N
alenahelenash

alenahelenash

Expert2022-01-23Added 556 answers

Step 1
Given that,
Weight of beam = 190 N
Here, The center of gravity is at its center
According to figure,
The angle is
sinθ=35
The horizontal component is
Tx=Tcos θ
The vertical component is
Ty=Tsinθ
a) We need to calculate the tension in the cable
Using formula of net torque acting on the pivot
τ=Fb×r+Fw×r'-Tsinθ×r'
Put the value into the formula
0=190×2+300×4-Tsinθ×4
Tsinθ×4=380+1200
T=1580×53×4
T=658.33N
Step 2
b) We need to calculate the horizontal components of the force exerted on the beam at the wal
Using formula of horizontal component
Fx=Tcosθ
Put the value into the formula
Fx=658.33×45
Fx=526.66N
c) We need to calculate the vertical components of the force exerted on the beam at the wall
Using formula of vertical component
Fy=Fb+Fw-Tsinθ
Put the value into the formula
Fy=190+300-658.33×35
Fy=95.002N

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