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3 years ago

» What is the value of s? S

Mathematics

2 answers:

bekas [8.4K]3 years ago

6 0

Answer:

s = 6

Step-by-step explanation:

soldier1979 [14.2K]3 years ago

4 0

Answer:

$\huge \boxed{s=6}$

$\boxed {\frak{Step \: by \: step \: explanation:}}$

$Given,3(s+4)=30$

$3s + 12 = 30$

$\boxed{Subtract 12 from \: both \: sides \: of \: the \: equation}$

$3s + 12 - 12 = 30 - 12$

$3s = 18$

$\boxed{Divide \: both \: sides \: of \: the \: equation \: by \: the \: same \: term}$

$\large{ \frac{3s}{3} = \frac{18}{3} }$

$\boxed{s = 6}$

$\frak{EtherealMistress}$

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Please red the answer below

Step-by-step explanation:

In order to determine the length of each cable you use the Newton second law for each component of the forces involved in the situation.

For the x component you have:

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T1: tension of the first cable = 1500N

T2: tension of the second cable = 800N

θ1: angle between the horizontal and the first cable

θ2: angle between the horizontal and the first cable

For the y component you have:

$T_1sin\theta_1+T_2sin\theta_2-W=0$ (2)

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You can squared both equations (1) and (2) and the sum the two equations:

$T_1^2cos^2\theta_1=T_2^2cos^2\theta_2\\\\T_1^2sin^2\theta_1=T_2^2sin^2\theta_2-2WT_2sin\theta_2+W^2$

Then, you sum the equations:

$T_1^2(cos^2\theta_1+sin^2\theta_1)=T_2^2(sin^2\theta_2+cos^2\theta_2)-2Wsin\theta_2+W^2$ (3)

Next, you use the following identity:

$sin^2\theta+cos^2\theta=1$

and you obtain in the equation (3):

$T_1^2=T_2^2-2WT_2sin\theta_2+W^2\\\\sin\theta_2=\frac{T_2^2-T_1^2+W^2}{2WT_2}=\frac{(800N)^2-(1500)^2+(1372N)^2}{2(800N)(1372N)}=0.066\\\\\theta_2=sin^{-1}0.066=27.23\°$

With this values you can calculate the value of the another angle, by using the equation (1):

$\theta_1=cos^{-1}(\frac{T_2cos(27.23\°)}{T_1})=cos^{-1}(\frac{(800N)(cos27.23\°)}{1500N})\\\\\theta_1=61.69\°$

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$l_1cos\theta_1=40-d\\\\l_2cos\theta_2=d\\\\$

d: distance to the right side of the field

By using the cosine law you can fins a system of equation and then you can calculate the values of l1 and l2.

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