Question

after an unfortunate accident occurred at a local warehouse,
you were contracted to determine the cause. a jib crane
collapsed and injured a worker. an image of this type of
crane is shown in the figure.
the horizontal steel beam had a mass of 91.90 kg per meter of
length, and the tension in the cable was ( t = 12170 mathrm{~n} ). the
crane was rated for a maximum load of 500 kg. the
acceleration due to gravity is ( g = 9.810 mathrm{~m} / mathrm{s}^{2} ).
if ( d = 5.580 mathrm{~m}, s = 0.486 mathrm{~m}, x = 1.300 mathrm{~m} ), and ( h = 1.890 mathrm{~m} ),
what was the magnitude of ( w_{l} ) (the load on the crane) before
the collapse?

Explanation:

Step1: Calculate the length of the beam and its mass

The length of the horizontal beam is \( L = d + s = 5.580\,\text{m}+ 0.486\,\text{m}=6.066\,\text{m} \).
The mass of the beam is \( m_{\text{beam}} = 91.90\,\text{kg/m} \times 6.066\,\text{m}\approx 557.47\,\text{kg} \).

Step2: Find the angle \(\theta\) using trigonometry

The vertical side for the cable triangle is \( h = 1.890\,\text{m} \), and the horizontal side is \( d + s - s = d = 5.580\,\text{m} \)? Wait, no—wait, the cable's triangle: the vertical component is \( h \), and the horizontal component is \( d + s - s = d \)? Wait, no, the cable is from the top of the vertical pole to the end of the beam (plus \( s \)). Wait, the horizontal distance for the cable is \( d + s \), and vertical distance is \( h \). Wait, no, the figure: the vertical pole has height \( h \), the beam extends \( d \) from the pole, then \( s \) more? Wait, the cable is attached to the top of the pole (height \( h \)) and to the beam at a distance \( d + s \) from the pole? Wait, the horizontal distance between the top of the pole and the cable attachment on the beam is \( d + s \), and vertical distance is \( h \). So the angle \(\theta\) has \(\sin\theta=\frac{h}{\sqrt{(d + s)^2 + h^2}}\)? Wait, no, wait: the horizontal segment from the pole to the cable attachment is \( d + s \), and vertical is \( h \). Wait, no, the problem says \( d = 5.580\,\text{m} \), \( s = 0.486\,\text{m} \), \( h = 1.890\,\text{m} \). Wait, actually, the cable's horizontal component is \( d + s \), and vertical component is \( h \)? Wait, no, looking at the figure: the vertical pole, the beam goes out \( d \), then \( s \) more, and the cable is from the top of the pole (height \( h \)) to the beam at the \( s \) end. So the horizontal distance between the top of the pole and the cable attachment is \( d + s \), and vertical distance is \( h \). Wait, but the torque is about point \( P \) (the pivot at the pole). So the tension \( T \) has a vertical component \( T\sin\theta \) and horizontal component \( T\cos\theta \), but for torque about \( P \), the horizontal component doesn't contribute (since its lever arm is zero). The vertical component of \( T \) provides a counter - torque, and the beam's weight and the load \( W_L \) provide clockwise torques.

Wait, no, torque about point \( P \) (the pivot where the beam meets the vertical pole). The tension \( T \) is applied at a distance \( d + s \) from \( P \), at an angle \(\theta\). The torque due to \( T \) is \( T\sin\theta\times(d + s) \) (since the perpendicular component of \( T \) to the beam is \( T\sin\theta \), and the lever arm is \( d + s \)). The torque due to the beam's weight: the beam's center of mass is at \( \frac{d + s}{2} \) from \( P \), so torque is \( m_{\text{beam}}g\times\frac{d + s}{2} \) (clockwise). The torque due to \( W_L \) is \( W_L\times x \) (clockwise, since \( x \) is the distance from \( P \) to the load). At equilibrium, the counter - torque (from \( T \)) equals the sum of clockwise torques.

Wait, let's re - express:

Torque about \( P \):

\( \sum\tau = 0 \)

Torque from \( T \): \( T\sin\theta\times(d + s) \) (counter - clockwise)

Torque from beam: \( m_{\text{beam}}g\times\frac{d + s}{2} \) (clockwise)

Torque from load: \( W_L\times x \) (clockwise)

So:

\( T\sin\theta\times(d + s)=m_{\text{beam}}g\times\frac{d + s}{2}+W_L\times x \)

We can solve for \( W_L \):

\( W_L=\frac{T\sin\theta\times(d + s)-m_{\text{beam}}g\times\frac{d + s}{2}}{x} \)

Now, find \( \sin\theta \): the vertical side is \( h = 1.890\,\text{m} \), and the hypotenuse of the cable's triangle is \( \sqrt{(d + s)^2 + h^2} \). Wait, no: the cable is from the top of the pole (height \( h \)) to the beam at a horizontal distance of \( d + s \) from the pole. So the length of the cable is \( \sqrt{(d + s)^2 + h^2} \), and \( \sin\theta=\frac{h}{\sqrt{(d + s)^2 + h^2}} \)? Wait, no, \( \sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{h}{\sqrt{(d + s)^2 + h^2}} \), and \( \cos\theta=\frac{d + s}{\sqrt{(d + s)^2 + h^2}} \). But in the torque, the perpendicular component of \( T \) to the beam is \( T\sin\theta \) (since the beam is horizontal, the angle between \( T \) and the vertical is related, but actually, the angle between \( T \) and the horizontal is \(\theta\), so the vertical component is \( T\sin\theta \) and horizontal is \( T\cos\theta \). The torque due to \( T \) about \( P \) is the vertical component times the horizontal distance from \( P \) to the cable attachment, which is \( d + s \). So yes, \( \tau_T = T\sin\theta\times(d + s) \).

First, calculate \( d + s = 5.580+0.486 = 6.066\,\text{m} \)

Calculate \( \sin\theta \):

\( \sin\theta=\frac{h}{\sqrt{(d + s)^2 + h^2}}=\frac{1.890}{\sqrt{(6.066)^2+(1.890)^2}} \)

First, compute denominator: \( (6.066)^2+(1.890)^2 = 36.796 + 3.572 = 40.368 \), square root is \( \sqrt{40.368}\approx6.354\,\text{m} \)

So \( \sin\theta=\frac{1.890}{6.354}\approx0.2975 \)

Now, compute the torque from \( T \): \( T\sin\theta\times(d + s)=12170\,\text{N}\times0.2975\times6.066\,\text{m} \)

First, \( 12170\times0.2975\approx12170\times0.3 - 12170\times0.0025 = 3651 - 30.425 = 3620.575\,\text{N} \)

Then, \( 3620.575\times6.066\approx3620.575\times6 + 3620.575\times0.066\approx21723.45+238.958\approx21962.41\,\text{N·m} \)

Torque from beam: \( m_{\text{beam}}g\times\frac{d + s}{2} \), \( m_{\text{beam}} = 91.90\,\text{kg/m}\times6.066\,\text{m}\approx557.47\,\text{kg} \)

\( \frac{d + s}{2}=\frac{6.066}{2}=3.033\,\text{m} \)

Torque from beam: \( 557.47\,\text{kg}\times9.810\,\text{m/s}^2\times3.033\,\text{m} \)

\( 557.47\times9.810\approx5468.88\,\text{N} \)

\( 5468.88\times3.033\approx5468.88\times3+5468.88\times0.033\approx16406.64 + 180.473\approx16587.11\,\text{N·m} \)

Now, set up the torque equation:

\( T\sin\theta(d + s)=m_{\text{beam}}g\frac{d + s}{2}+W_Lx \)

Solve for \( W_L \):

\( W_L=\frac{T\sin\theta(d + s)-m_{\text{beam}}g\frac{d + s}{2}}{x} \)

Plug in the numbers:

Numerator: \( 21962.41 - 16587.11 = 5375.3\,\text{N·m} \)

\( x = 1.300\,\text{m} \)

\( W_L=\frac{5375.3}{1.300}\approx4134.85\,\text{N} \)? Wait, that can't be right, because the crane is rated for 500 kg, which is \( 500\times9.81 = 4905\,\text{N} \). Wait, maybe I made a mistake in the angle. Wait, maybe the horizontal distance for the cable is \( d \), not \( d + s \)? Wait, looking back at the figure: the beam has a length from the pole of \( d \), then \( s \) more? Wait, the problem says \( d = 5.580\,\text{m} \), \( s = 0.486\,\text{m} \), so the cable is attached at \( d + s \) from the pole? Or is \( s \) the overhang? Wait, maybe the horizontal distance between the top of the pole and the cable attachment is \( d \), and \( s \) is something else. Wait, no, the problem's figure: the beam is horizontal, with \( P \) at the pole, then \( d \) to a point, then \( s \) to the cable attachment, and \( x \) from the cable attachment to the load? Wait, no, the load is at \( x \) from \( P \). Wait, maybe the cable is attached at a distance \( d \) from \( P \), and \( s \) is a small segment? Wait, no, the problem says \( d = 5.580\,\text{m} \), \( s = 0.486\,\text{m} \), \( h = 1.890\,\text{m} \). Wait, maybe the horizontal component of the cable is \( d \), and vertical is \( h \), so the angle \(\theta\) has \( \tan\theta=\frac{h}{d} \)? Wait, that would make more sense. Let's re - calculate \(\sin\theta\) with horizontal distance \( d \), not \( d + s \).

Ah! That's the mistake. The cable is attached to the beam at a distance \( d \) from \( P \), and \( s \) is a small segment? Wait, no, the figure: the beam extends \( d \) from \( P \), then \( s \) to the end, and the cable is from the top of the pole (height \( h \)) to the beam at \( d \) from \( P \)? No, the problem says \( d = 5.580\,\text{m} \), \( s = 0.486\,\text{m} \), so the total length from \( P \) to the end of the beam is \( d + s \), and the cable is attached at the end? No, the load is at \( x = 1.300\,\text{m} \) from \( P \), so \( x \) is less than \( d \). Wait, maybe the cable is attached at a distance \( d \) from \( P \), and the horizontal distance between the top of the pole and the cable attachment is \( d \), vertical distance \( h \). So the angle \(\theta\) has \( \sin\theta=\frac{h}{\sqrt{d^2 + h^2}} \), and \( \cos\theta=\frac{d}{\sqrt{d^2 + h^2}} \). Let's try that.

So \( d = 5.580\,\text{m} \), \( h = 1.890\,\text{m} \)

Denominator: \( \sqrt{(5.580)^2+(1.890)^2}=\sqrt{31.136 + 3.572}=\sqrt{34.708}\approx5.891\,\text{m} \)

\( \sin\theta=\frac{1.890}{5.891}\approx0.3208 \)

Now, torque from \( T \): the cable is attached at distance \( d \) from \( P \)? No, the load is at \( x = 1.300\,\text{m} \) from \( P \), and the cable is attached at the end of the beam ( \( d + s \) from \( P \))? Wait, the problem says "the horizontal steel beam had a mass of 91.90 kg per meter of length", so the beam's length is \( L = d + s \), as before. The center of mass is at \( L/2=(d + s)/2 \). The load is at \( x = 1.300\,\text{m} \) from \( P \). The cable is attached at the end of the beam ( \( d + s \) from \( P \)), with tension \( T \), making an angle \(\theta\) with the beam. So the torque due to \( T \) is \( T\sin\theta\times(d + s) \) (counter - clockwise), torque due to beam is \( m_{\text{beam}}g\times\frac{d + s}{2} \) (clockwise), torque due to load is \( W_L\times x \) (clockwise).

Let's recalculate with \( d + s = 6.066\,\text{m} \), \( h = 1.890\,\text{m} \), so the cable's vertical component is \( h \), horizontal component is \( d + s \)? No, that can't be, because \( h = 1.890\) and \( d + s = 6.066 \), so the triangle is tall and thin? No, \( h \) is the height of the pole, so the cable goes from the top of the pole (height \( h \)) to the end of the beam (horizontal distance \( d + s \) from the pole). So the angle \(\theta\) is between the cable and the beam (horizontal), so \( \tan\theta=\frac{h}{d + s} \), so \( \sin\theta=\frac{h}{\sqrt{(d + s)^2 + h^2}} \), as I first thought, but maybe my calculation of \( m_{\text{beam}} \) was wrong. Wait, \( m_{\text{beam}} = 91.90\,\text{kg/m}\times L \), where \( L \) is the length of the beam. The beam's length is \( d + s = 5.580+0.486 = 6.066\,\text{m} \), so \( m_{\text{beam}} = 91.90\times6.066\approx91.90\times6 + 91.90\times0.066 = 551.4+6.0654 = 557.4654\,\text{kg} \), that's correct.

Torque from \( T \): \( T\sin\theta\times(d + s) \), \( \sin\theta=\frac{h}{\sqrt{(d + s)^2 + h^2}}=\frac{1.890}{\sqrt{6.066^2 + 1.890^2}}=\frac{1.890}{\sqrt{36.796 + 3.572}}=\frac{1.890}{\sqrt{40.368}}\approx\frac{1.890}{6.354}\approx0.2975 \)

So \( T\sin\theta\times(d + s)=12170\times0.2975\times6.066 \)

\( 12170\times0.2975 = 12170\times\frac{1.890}{\sqrt{6.066^2 + 1.890^2}} \)? Wait, no, \( \sin\theta=\frac{h}{\text{cable length}} \),

Answer:

Step1: Calculate the length of the beam and its mass

The length of the horizontal beam is \( L = d + s = 5.580\,\text{m}+ 0.486\,\text{m}=6.066\,\text{m} \).
The mass of the beam is \( m_{\text{beam}} = 91.90\,\text{kg/m} \times 6.066\,\text{m}\approx 557.47\,\text{kg} \).

Step2: Find the angle \(\theta\) using trigonometry

The vertical side for the cable triangle is \( h = 1.890\,\text{m} \), and the horizontal side is \( d + s - s = d = 5.580\,\text{m} \)? Wait, no—wait, the cable's triangle: the vertical component is \( h \), and the horizontal component is \( d + s - s = d \)? Wait, no, the cable is from the top of the vertical pole to the end of the beam (plus \( s \)). Wait, the horizontal distance for the cable is \( d + s \), and vertical distance is \( h \). Wait, no, the figure: the vertical pole has height \( h \), the beam extends \( d \) from the pole, then \( s \) more? Wait, the cable is attached to the top of the pole (height \( h \)) and to the beam at a distance \( d + s \) from the pole? Wait, the horizontal distance between the top of the pole and the cable attachment on the beam is \( d + s \), and vertical distance is \( h \). So the angle \(\theta\) has \(\sin\theta=\frac{h}{\sqrt{(d + s)^2 + h^2}}\)? Wait, no, wait: the horizontal segment from the pole to the cable attachment is \( d + s \), and vertical is \( h \). Wait, no, the problem says \( d = 5.580\,\text{m} \), \( s = 0.486\,\text{m} \), \( h = 1.890\,\text{m} \). Wait, actually, the cable's horizontal component is \( d + s \), and vertical component is \( h \)? Wait, no, looking at the figure: the vertical pole, the beam goes out \( d \), then \( s \) more, and the cable is from the top of the pole (height \( h \)) to the beam at the \( s \) end. So the horizontal distance between the top of the pole and the cable attachment is \( d + s \), and vertical distance is \( h \). Wait, but the torque is about point \( P \) (the pivot at the pole). So the tension \( T \) has a vertical component \( T\sin\theta \) and horizontal component \( T\cos\theta \), but for torque about \( P \), the horizontal component doesn't contribute (since its lever arm is zero). The vertical component of \( T \) provides a counter - torque, and the beam's weight and the load \( W_L \) provide clockwise torques.

Wait, no, torque about point \( P \) (the pivot where the beam meets the vertical pole). The tension \( T \) is applied at a distance \( d + s \) from \( P \), at an angle \(\theta\). The torque due to \( T \) is \( T\sin\theta\times(d + s) \) (since the perpendicular component of \( T \) to the beam is \( T\sin\theta \), and the lever arm is \( d + s \)). The torque due to the beam's weight: the beam's center of mass is at \( \frac{d + s}{2} \) from \( P \), so torque is \( m_{\text{beam}}g\times\frac{d + s}{2} \) (clockwise). The torque due to \( W_L \) is \( W_L\times x \) (clockwise, since \( x \) is the distance from \( P \) to the load). At equilibrium, the counter - torque (from \( T \)) equals the sum of clockwise torques.

Wait, let's re - express:

Torque about \( P \):

\( \sum\tau = 0 \)

Torque from \( T \): \( T\sin\theta\times(d + s) \) (counter - clockwise)

Torque from beam: \( m_{\text{beam}}g\times\frac{d + s}{2} \) (clockwise)

Torque from load: \( W_L\times x \) (clockwise)

So:

\( T\sin\theta\times(d + s)=m_{\text{beam}}g\times\frac{d + s}{2}+W_L\times x \)

We can solve for \( W_L \):

\( W_L=\frac{T\sin\theta\times(d + s)-m_{\text{beam}}g\times\frac{d + s}{2}}{x} \)

Now, find \( \sin\theta \): the vertical side is \( h = 1.890\,\text{m} \), and the hypotenuse of the cable's triangle is \( \sqrt{(d + s)^2 + h^2} \). Wait, no: the cable is from the top of the pole (height \( h \)) to the beam at a horizontal distance of \( d + s \) from the pole. So the length of the cable is \( \sqrt{(d + s)^2 + h^2} \), and \( \sin\theta=\frac{h}{\sqrt{(d + s)^2 + h^2}} \)? Wait, no, \( \sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{h}{\sqrt{(d + s)^2 + h^2}} \), and \( \cos\theta=\frac{d + s}{\sqrt{(d + s)^2 + h^2}} \). But in the torque, the perpendicular component of \( T \) to the beam is \( T\sin\theta \) (since the beam is horizontal, the angle between \( T \) and the vertical is related, but actually, the angle between \( T \) and the horizontal is \(\theta\), so the vertical component is \( T\sin\theta \) and horizontal is \( T\cos\theta \). The torque due to \( T \) about \( P \) is the vertical component times the horizontal distance from \( P \) to the cable attachment, which is \( d + s \). So yes, \( \tau_T = T\sin\theta\times(d + s) \).

First, calculate \( d + s = 5.580+0.486 = 6.066\,\text{m} \)

Calculate \( \sin\theta \):

\( \sin\theta=\frac{h}{\sqrt{(d + s)^2 + h^2}}=\frac{1.890}{\sqrt{(6.066)^2+(1.890)^2}} \)

First, compute denominator: \( (6.066)^2+(1.890)^2 = 36.796 + 3.572 = 40.368 \), square root is \( \sqrt{40.368}\approx6.354\,\text{m} \)

So \( \sin\theta=\frac{1.890}{6.354}\approx0.2975 \)

Now, compute the torque from \( T \): \( T\sin\theta\times(d + s)=12170\,\text{N}\times0.2975\times6.066\,\text{m} \)

First, \( 12170\times0.2975\approx12170\times0.3 - 12170\times0.0025 = 3651 - 30.425 = 3620.575\,\text{N} \)

Then, \( 3620.575\times6.066\approx3620.575\times6 + 3620.575\times0.066\approx21723.45+238.958\approx21962.41\,\text{N·m} \)

Torque from beam: \( m_{\text{beam}}g\times\frac{d + s}{2} \), \( m_{\text{beam}} = 91.90\,\text{kg/m}\times6.066\,\text{m}\approx557.47\,\text{kg} \)

\( \frac{d + s}{2}=\frac{6.066}{2}=3.033\,\text{m} \)

Torque from beam: \( 557.47\,\text{kg}\times9.810\,\text{m/s}^2\times3.033\,\text{m} \)

\( 557.47\times9.810\approx5468.88\,\text{N} \)

\( 5468.88\times3.033\approx5468.88\times3+5468.88\times0.033\approx16406.64 + 180.473\approx16587.11\,\text{N·m} \)

Now, set up the torque equation:

\( T\sin\theta(d + s)=m_{\text{beam}}g\frac{d + s}{2}+W_Lx \)

Solve for \( W_L \):

\( W_L=\frac{T\sin\theta(d + s)-m_{\text{beam}}g\frac{d + s}{2}}{x} \)

Plug in the numbers:

Numerator: \( 21962.41 - 16587.11 = 5375.3\,\text{N·m} \)

\( x = 1.300\,\text{m} \)

\( W_L=\frac{5375.3}{1.300}\approx4134.85\,\text{N} \)? Wait, that can't be right, because the crane is rated for 500 kg, which is \( 500\times9.81 = 4905\,\text{N} \). Wait, maybe I made a mistake in the angle. Wait, maybe the horizontal distance for the cable is \( d \), not \( d + s \)? Wait, looking back at the figure: the beam has a length from the pole of \( d \), then \( s \) more? Wait, the problem says \( d = 5.580\,\text{m} \), \( s = 0.486\,\text{m} \), so the cable is attached at \( d + s \) from the pole? Or is \( s \) the overhang? Wait, maybe the horizontal distance between the top of the pole and the cable attachment is \( d \), and \( s \) is something else. Wait, no, the problem's figure: the beam is horizontal, with \( P \) at the pole, then \( d \) to a point, then \( s \) to the cable attachment, and \( x \) from the cable attachment to the load? Wait, no, the load is at \( x \) from \( P \). Wait, maybe the cable is attached at a distance \( d \) from \( P \), and \( s \) is a small segment? Wait, no, the problem says \( d = 5.580\,\text{m} \), \( s = 0.486\,\text{m} \), \( h = 1.890\,\text{m} \). Wait, maybe the horizontal component of the cable is \( d \), and vertical is \( h \), so the angle \(\theta\) has \( \tan\theta=\frac{h}{d} \)? Wait, that would make more sense. Let's re - calculate \(\sin\theta\) with horizontal distance \( d \), not \( d + s \).

Ah! That's the mistake. The cable is attached to the beam at a distance \( d \) from \( P \), and \( s \) is a small segment? Wait, no, the figure: the beam extends \( d \) from \( P \), then \( s \) to the end, and the cable is from the top of the pole (height \( h \)) to the beam at \( d \) from \( P \)? No, the problem says \( d = 5.580\,\text{m} \), \( s = 0.486\,\text{m} \), so the total length from \( P \) to the end of the beam is \( d + s \), and the cable is attached at the end? No, the load is at \( x = 1.300\,\text{m} \) from \( P \), so \( x \) is less than \( d \). Wait, maybe the cable is attached at a distance \( d \) from \( P \), and the horizontal distance between the top of the pole and the cable attachment is \( d \), vertical distance \( h \). So the angle \(\theta\) has \( \sin\theta=\frac{h}{\sqrt{d^2 + h^2}} \), and \( \cos\theta=\frac{d}{\sqrt{d^2 + h^2}} \). Let's try that.

So \( d = 5.580\,\text{m} \), \( h = 1.890\,\text{m} \)

Denominator: \( \sqrt{(5.580)^2+(1.890)^2}=\sqrt{31.136 + 3.572}=\sqrt{34.708}\approx5.891\,\text{m} \)

\( \sin\theta=\frac{1.890}{5.891}\approx0.3208 \)

Now, torque from \( T \): the cable is attached at distance \( d \) from \( P \)? No, the load is at \( x = 1.300\,\text{m} \) from \( P \), and the cable is attached at the end of the beam ( \( d + s \) from \( P \))? Wait, the problem says "the horizontal steel beam had a mass of 91.90 kg per meter of length", so the beam's length is \( L = d + s \), as before. The center of mass is at \( L/2=(d + s)/2 \). The load is at \( x = 1.300\,\text{m} \) from \( P \). The cable is attached at the end of the beam ( \( d + s \) from \( P \)), with tension \( T \), making an angle \(\theta\) with the beam. So the torque due to \( T \) is \( T\sin\theta\times(d + s) \) (counter - clockwise), torque due to beam is \( m_{\text{beam}}g\times\frac{d + s}{2} \) (clockwise), torque due to load is \( W_L\times x \) (clockwise).

Let's recalculate with \( d + s = 6.066\,\text{m} \), \( h = 1.890\,\text{m} \), so the cable's vertical component is \( h \), horizontal component is \( d + s \)? No, that can't be, because \( h = 1.890\) and \( d + s = 6.066 \), so the triangle is tall and thin? No, \( h \) is the height of the pole, so the cable goes from the top of the pole (height \( h \)) to the end of the beam (horizontal distance \( d + s \) from the pole). So the angle \(\theta\) is between the cable and the beam (horizontal), so \( \tan\theta=\frac{h}{d + s} \), so \( \sin\theta=\frac{h}{\sqrt{(d + s)^2 + h^2}} \), as I first thought, but maybe my calculation of \( m_{\text{beam}} \) was wrong. Wait, \( m_{\text{beam}} = 91.90\,\text{kg/m}\times L \), where \( L \) is the length of the beam. The beam's length is \( d + s = 5.580+0.486 = 6.066\,\text{m} \), so \( m_{\text{beam}} = 91.90\times6.066\approx91.90\times6 + 91.90\times0.066 = 551.4+6.0654 = 557.4654\,\text{kg} \), that's correct.

Torque from \( T \): \( T\sin\theta\times(d + s) \), \( \sin\theta=\frac{h}{\sqrt{(d + s)^2 + h^2}}=\frac{1.890}{\sqrt{6.066^2 + 1.890^2}}=\frac{1.890}{\sqrt{36.796 + 3.572}}=\frac{1.890}{\sqrt{40.368}}\approx\frac{1.890}{6.354}\approx0.2975 \)

So \( T\sin\theta\times(d + s)=12170\times0.2975\times6.066 \)

\( 12170\times0.2975 = 12170\times\frac{1.890}{\sqrt{6.066^2 + 1.890^2}} \)? Wait, no, \( \sin\theta=\frac{h}{\text{cable length}} \),