A boat owner pulls her boat into the dock shown, where

thereare six capstans to which to tie the boat. She has three

ropes. Shecan tie the boat from the boat’s center (A) to any of the

capstans(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

aretied such that

.

The following notation is used in

thisproblem: When a question refers to, for example, , this

quantity is taken to mean the force acting on theboat due to the

rope running to A from B, while is the

magnitude of that force.

The following notation is used in

thisproblem: When a question refers to, for example, , this

quantity is taken to mean the force acting on theboat due to the

rope running to A from B, while is the

magnitude of that force. What is the

magnitude of the force provided by thethird rope, in terms of

?

A. FABCos(θ)

B. 2FABCos(θ)

C. 2FABSin(θ)

D.FABSin(θ)

I think its C. Am i right?

What is the

magnitude of the force provided by thethird rope, in terms of

? What is the

magnitude of the force provided by thethird rope, in terms of

? A. FABCos(θ) AB B. 2FABCos(θ) AB C. 2FABSin(θ) AB D.FABSin(θ) AB I think its C. Am i right?

General guidance

Concepts and reason

The concepts required to solve this problem are equilibrium force condition and vector components.

Initially, use the component equation of a force to resolve the forces along the axes. Then, determine the net force along the horizontal direction. Finally, use the equilibrium force condition to solve for the magnitude of the force by third rope.

Fundamentals

The equilibrium force condition gives that the sum of all the forces on an object is zero when it is not accelerating. This is expressed as,

Here, is sum of all the forces on the object.

The components of a force are as follows:

Here, is the component of the force, is the component of the force, is the magnitude of force, and is the angle between the axis and force in counterclockwise direction.

Sign Convention is as follows:

Take all the forces in upward or towards right as positive and all the forces in downward or towards left as negative.

Step-by-step

Step 1 of 2

Draw the diagram of the boat and ropes showing the direction of the forces. ,, and are the forces acting on the boat due to the rope running to A from B, to A from D, and to A from F. The components of the force along and components are and . The components of the force along and components are and . The components of the force along and components are and .

Use the component equation of a force and substitute for , and for in the equation .

Use the component equation of a force and substitute for , and for in the equation .

In the above diagram, a boat owner pulls her boat into the dock there are six bollards to which to tie the boat. Owner has three ropes and can tie the boat from the boat’s center A to any of the bollards. One rope runs to A from B, another to A from D, and a final rope from A to F.

The component of the force along the horizontal direction is not given as, . The component of the force along the horizontal direction is given as, . Here, is the angle between the force and the horizontal component of the force.

Use the equilibrium force condition to solve for the magnitude of force provided by third rope.

Step 2 of 2

Use the equilibrium force condition along axis.

Three forces act on the boat while it is stationary, and towards left so negative and towards right so positive. Substitute for in the above equation .

Substitute for , and for in the above equation.

Since, substitute for in the above equation.

The magnitude of force provided by the third rope in terms of is . So, the options are not correct.

The magnitude of force provided by the third rope in terms of is .

Three forces act on the boat while it is stationary, and towards left so negative and towards right so positive. So, the net force acting on the boat along the horizontal direction is,

The equilibrium force condition gives that the sum of all the forces on an object is zero when it is not accelerating is zero. As the boat is not accelerating along the horizontal direction. The net force along the horizontal direction is zero. That is,

The net force along the horizontal direction is not given as, . As the forces and are acting towards and towards right, so the net force along the horizontal direction is given as,

Answer

The magnitude of force provided by the third rope in terms of is .

Answer only

The magnitude of force provided by the third rope in terms of is .

ΣF = 0

F = F cos F, = F sin

Boat Figure 1: The boat and rope diagram with all the forces and its components.

F = F cos

Fare = Fab cose

F = F cos

Fadx = Fan cos

F = F sin

F = F cos

ΣF = 0

FF – FABE – FADE

ΣF = 0

Far – FABr – Fadx = 0 Far = FABr +FADE

FAB cos

Fap cos

Far = FAB cosO+Fap cos

FFD

Far = FAB cos 0 + FAB cos = 2F cos

2FAB cos

FAB cos 0, 2F 3 sin 0 and Fall sin

F=FF-FAB – Fax

ΣF = 0 Fx – Fikr – Fxdx = 0

EF = Fxp+Fxxx + FADE

F=FF-FAB – Fax

2FAB cos

2FAB cos

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