Question

A boat owner pulls her boat into the dock shown, where there are six bollards to which to tie the boat. She has three ropes. She can tie the boat from the boat's center (A) to any of the bollards (B through G) along the dotted arrows shown. Suppose the owner has tied three ropes: one rope runs to A from B, another to A from D, and a final rope from A to F. The ropes are tied such that FAB=FAD.

The following notation is used in this problem: When a question refers to, for example, F⃗ AB, this quantity is taken to mean the force acting on the boat due to the rope running to A from B, while FAB is the magnitude of that force. Part A What is the magnitude of the force provided by the third rope, in terms of θ? What is the magnitude of the force provided by the third rope, in terms of ? FABcos(θ) 2FABcos(θ) 2FABsin(θ) FABsin(θ)

Answer 1

Concepts and Reason

The concept required to solve this question is the Newton’s second law of motion.

First, write the expression for the net force along the horizontal direction by using the Newton’s second law of motion. Finally, calculate the force along the third rope by using the relation between the force acting on the boat due to rope running A from B and the force acting on the boat due to rope running A from D.

Fundamentals

A force is that which changes or tends to change the state of rest or motion of a body. It is vector quantity. A vector quantity is a quantity that has both direction and magnitude. Force is a pull or push upon an object resulting from the objects interaction with another object. Boat owner has three ropes. Owner can tie the boat from the boat's center to any of the bollards. Calculate the magnitude of force provided by third rope.

The general expression of force is,

$F = ma$

Here, $m$ is the mass and $a$ is the acceleration.

(A)

Let ${F_{{\rm{AB}}}}$ be force acting on the boat due to rope running A from B. The force along the rope ${\rm{AB}}$ has two components. Horizontal component is ${F_{{\rm{AB}}}}\cos \theta$ and vertical component is ${F_{{\rm{AB}}}}\sin \theta$ .

Let ${F_{{\rm{AD}}}}$ be the force acting on the boat due to rope running A from D. The force along the rope ${\rm{AD}}$ has two components. Horizontal component is ${F_{{\rm{AD}}}}\cos \theta$ and vertical component is ${F_{{\rm{AD}}}}\sin \theta$ .

The free-body diagram for the forces acting on the boat is shown in the below figure.

The net force along the horizontal direction is,

${F_{{\rm{AB}}}}\cos \theta + {F_{{\rm{AD}}}}\cos \theta = {F_{{\rm{AF}}}}$

Given that, the force acting on the boat due to rope running A from B is equal to the force acting on the boat due to rope running A from D that is ${F_{{\rm{AB}}}} = {F_{{\rm{AD}}}}$ .

Substitute ${F_{{\rm{AB}}}}$ for ${F_{{\rm{AD}}}}$ in above equation.

$\begin{array}{c}\\{F_{{\rm{AB}}}}\cos \theta + {F_{{\rm{AB}}}}\cos \theta = {F_{{\rm{AF}}}}\\\\{F_{{\rm{AF}}}} = 2{F_{{\rm{AB}}}}\cos \theta \\\end{array}$

Ans: Part A

Magnitude of force along third rope is $2{F_{{\rm{AB}}}}\cos \theta$ .