Physics, 23.03.2020 16:56 Vanesssajohn2962

A large ship is being towed by two tugs. The larger tug exerts a force which is 20 percent greater than the smaller tug, and at an angle of 35 degrees south of west. Which direction must the smaller tug pull to ensure that the ship travels due west?

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Answer from: azibur3191

direction is 43.495 degrees south of west.

Explanation:

given data

larger tug exert force = 20 percent greater than smaller tug

angle = 35 degrees south of west

solution

The smaller tug must pull in south west direction at angle that from north - south component of 2 force that cancel each other out

we take smaller tug is pulling with a force F

Smaller tug express as = F sin(α)

and

Larger tug is (F + 0.20 F) sin(35 degrees) < 20% larger than smaller tugboat and 0 degree direction is directly west

so here we can say

Smaller tug + Larger tug = Fsinα + (F + 0.20 F) sin(35 ) = 0

here α angle is going to be south of west

so it will be a negative angle.

and

now we Divide F of the entire equation so

it will be as

sinα + (1 + 0.20 ) sin(35 ) = sinα + 0.6596 = 0

sinα + 0.6883 = 0

sinα = -0.6883

α = -43.495 degree

so direction is 43.495 degrees south of west.

Answer from: Quest

here we know that ball will drop down by total distance 3.2 m

so here we know that

$v_{yi} = 0$

$\delta y = - 3.2 m$

now using kinematics

$\delta y = v_{yi} t + \frac{1}{2}at^2$

$-3.2 = 0 -\frac{1}{2}(9.8) t^2$

$t = 0.81 s$

now in the same time it will cover horizontal distance of 0.75 m

$\delta x = 0.75m$

so speed of the ball will be

$v = \frac{delta x }{\delta t}$

$v = \frac{0.75}{0.81} = 0.93 m/s$

Answer from: Quest

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