Question

If it takes 35 s for the 50-Mg tugboat to increase its speed uniformly to 25 km/h, starting from rest, determine the force of the rope on the tugboat. The propeller provides the propulsion force F which gives the tugboat forward motion, whereas the barge moves freely. Also, determine F acting on the tugboat. The barge has a mass of 75 Mg.

Answer 1

- the force of the rope on the tugboat is 14.87 kN

- Force acting on the tugboat is 24.79 kN

Explanation:

Given the data in the question;

tugboat increases its speed uniformly to 25 km/h

v₁ = 25 km/h = (25 × 1000) / ( 1 × 60min × 60sec )

= 25000m / 3600s

= 6.94 m/s

Now, lets determine the force on the rope using the following relation;

T =
v₁ / t₁

is mass of barge( 75 Megagram = 75 × 10³ Kilogram ), time t₁ is 35 s and v₁ is 6.94 m/s

so we substitute

T = [(75 × 10³) × 6.94 ] / 35

T = 520500 / 35

T = 14871.43 N

T = 14871.43 / 1000

T = 14.87 kN

Therefore, the force of the rope on the tugboat is 14.87 kN

Now, to determine F acting on the tugboat;

= (
+
)

we solve for F

F = (
+
)
/

where
is mass of tugboat (50 Megagram = 50 × 10³ Kilogram )

so we substitute

F = [( (50 × 10³) + (75 × 10³) )6.94] / 35

F = [ 125000 × 6.94 ) / 35

F = 867500 / 35

F = 24785.7 N

F = 24785.7 / 1000

F = 24.79 kN

Therefore, Force acting on the tugboat is 24.79 kN