Question

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.90×106 N, one at an angle 19.0 west of north, and the other at an angle 19.0 east of north, as they pull the tanker a distance 0.870 km toward the north.

What is the total work done by the two tugboats on the supertanker?

Express your answer in joules, to three significant figures.

Answer 1

Concepts and reason

The concepts required to answer the problem are resolution of forces and work done due to a force.

The direction of forces applied is given. The forces can be resolved into its components and the net force can be calculated. Also, as the displacement of the supertanker is given, the work done by the tugboats can be calculated.

Fundamentals

The forces are said to be balanced, when the net force along a specific direction is zero.

Resolution of forces is the splitting of a forces into its different components which are perpendicular to each other.

The Force F can be split into two components as shown

$F = {F_{\rm{x}}} + {F_{\rm{y}}}$

Here $F\,$ is the net force, ${F_{\rm{x}}}$ is the horizontal component of force, ${F_{\rm{y}}}$ is the vertical component of force

$F = F\cos \theta + F\sin \theta$

Here $F$ is the force and $\theta$ is the angle made by the force with $x$ axis.

Work is said to be done when a force acts on body and displaces it by some distance.

It is given by

$W.D = Fd\cos \theta$

Here, $F$ is the force, $d$ is the displacement and $\theta$ is the angle between force vector and displacement vector.

Its unit is Joules and is denoted by J.

The given forces can be resolved as follow

Since, the force’s component towards East is equal to the force’s components towards west, the force along East – West line will be balanced.

The other component of forces lies in the same direction, towards the North. Therefore, net force will be towards the north.

Net force is towards north is given by

${F_{{\rm{net}}}} = 2F\cos {19^0}$

Substitute $1.90 \times {10^6}{\rm{N}}$ for $F$

$\begin{array}{l}\\{F_{{\rm{net}}}} = 2\left( {1.90 \times {{10}^6}{\rm{N}}} \right)\cos {19^0}\\\\{F_{{\rm{net}}}} = 3.59 \times {10^6}{\rm{N}}\\\end{array}$

Work done is given by

$W.D = F.d\cos \theta$

Substitute $3.59 \times {10^6}{\rm{N}}$ for $F$ , $0.870 \times {10^3}{\rm{m}}$ for $d$ and ${0^0}$ for $\theta$

$W.D = \left( {3.59 \times {{10}^6}{\rm{N}}} \right)\left( {0.870 \times {{10}^3}{\rm{m}}} \right)\cos {0^0}$

$\begin{array}{l}\\W.D = \left( {3.59 \times {{10}^6}{\rm{N}}} \right)\left( {0.870 \times {{10}^3}{\rm{m}}} \right)\\\\W.D = 3.12 \times {10^9}{\rm{J}}\\\end{array}$

Ans:

The work done by the tugboats is $W.D = 3.12 \times {10^9}{\rm{J}}$