Question

How do i do this yall

Answer 1

The tension in the left-hand cable is 110 N.

We can solve this problem using the following steps:

Equations of equilibrium:

∑F_x = 0

`T_2` cos
`\theta_2` -
`T_1` cos
`\theta_1` = 0

∑F_y = 0

`T_2` sin
`\theta_2` +
`T_1` sin
`\theta_1` - W = 0

Solving for
`T_2`:

From the first equation, we can solve for
`T_1`:

`T_1` cos
`\theta_1` =
`T_2` cos
`\theta_2`

Substituting this into the second equation, we get:

`T_2` sin
`\theta_2` +
`T_2` cos
`\theta_2` / cos
`\theta_1` * sin
`\theta_1` - W = 0

`T_2` sin
`\theta_2` +
`T_2` cos
`\theta_2` tan
`\theta_1` - W = 0

We know the following values:

W = 624 N

`\theta_1` = 62°

`\theta_2` = 50°

We can also solve for cos
`\theta_2` using the Pythagorean theorem:

cos
`\theta_2` =
`√((1 - sin^2 \theta_2))`

cos
`\theta_2` = √(1 - 0.64)

cos
`\theta_2` = 0.766

Substituting all of these values into the equation above, we get:

`T_2` sin 50° +
`T_2` * 0.766 * tan 62° - 624 = 0

`T_2` = 110 N

Therefore, the tension in the left-hand cable is 110 N.