Answer:
(a). The tension in the cable is 658.33 N.
(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.
(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.
Explanation:
Given that,
Weight of beam= 190 N
Here, The center of gravity is at its center
According to figure,
The angle is
[tex]\sin\theta=\dfrac{3}{5}[/tex]
The horizontal component is
[tex]T_{x}=T\cos\theta[/tex]
The vertical component is
[tex]T_{y}=T\sin\theta[/tex]
(a). We need to calculate the tension in the cable
Using formula of net torque acting on the pivot
[tex]\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'[/tex]
Put the value into the formula
[tex]0=190\times2+300\times 4-T\sin\theta\times 4[/tex]
[tex]T\sin\theta\times 4=380+1200[/tex]
[tex]T=\dfrac{1580\times5}{3\times 4}[/tex]
[tex]T=658.33\ N[/tex]
(b). We need to calculate the horizontal components of the force exerted on the beam at the wall
Using formula of horizontal component
[tex]F_{x}=T\cos\theta[/tex]
Put the value into the formula
[tex]F_{x}=658.33\times\dfrac{4}{5}[/tex]
[tex]F_{x}=526.66\ N[/tex]
(c). We need to calculate the vertical components of the force exerted on the beam at the wall
Using formula of vertical component
[tex]F_{y}=F_{b}+F_{w}-T\sin\theta[/tex]
Put the value into the formula
[tex]F_{y}=190+300-658.33\times\dfrac{3}{5}[/tex]
[tex]F_{y}=95.002\ N[/tex]
Hence, (a). The tension in the cable is 658.33 N.
(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.
(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.
The tension in the cable is 658.33 N.
The horizontal and vertical components of the force exerted on the beam at the wall are 526.66 N and 95.002 N. respectively.
Given:
Weight of beam= 190 N
The horizontal component= Tcosθ
The vertical component= Tsinθ
To calculate the tension of the cable
Tsinθ x 4 = 380 + 1200
=658.33 N.
To find the horizontal component
=658.33 N x 4/5
= 526.66 N.
To find the vertical component
=190 + 300 -658.33 x 3/5
=95.002 N
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