A boat owner pulls her boat right into the dock displayed, where there aresix bollards to which to tie the boat. She has 3 ropes. She cantie the boat from the boat's center (A) to any type of of the bollards (Bvia G) alengthy the dotted arrows displayed.
Suppose the owner has actually tied 3 ropes: one rope runs to A fromB, one more to A from D, and a last rope from A to F. The ropes aretied such that FAB=FAD.
The following notation is provided in this problem: When a questiondescribes, for example, F⃗ AB, this amount is taken tomean the force acting on the watercraft due to the rope running to A fromB, while FAB is the magnitude of that pressure.
What is the magnitude of the force gave by the 3rd rope,in terms of θ?What is the magnitude of the pressure provided by the third rope, inregards to ?
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Concepts and also Reason
The concept compelled to resolve this question is the Newton’s second legislation of motion.
First, create the expression for the net force alengthy the horizontal direction by utilizing the Newton’s second law of movement. Finally, calculate the pressure along the 3rd rope by using the relation in between the force acting on the boat because of rope running A from B and also the pressure acting on the watercraft because of rope running A from D.
A pressure is that which changes or has a tendency to adjust the state of rest or motion of a body. It is vector quantity. A vector amount is a amount that has actually both direction and also magnitude. Force is a pull or push upon a things resulting from the objects interaction with an additional object. Boat owner has three ropes. Owner deserve to tie the boat from the boat's facility to any kind of of the bollards. Calculate the magnitude of force provided by third rope.
The general expression of pressure is,
F=maF = maF=ma
Here, mmm is the mass and also aaa is the acceleration.
Let FABF_ mABFAB be force acting on the watercraft because of rope running A from B. The pressure along the rope AB mABAB has 2 components. Horizontal component is FABcosθF_ mABcos heta FABcosθ and also vertical component is FABsinθF_ mABsin heta FABsinθ .
Let FADF_ mADFAD be the pressure acting on the watercraft because of rope running A from D. The pressure alengthy the rope AD mADAD has actually two components. Horizontal component is FADcosθF_ mADcos heta FADcosθ and vertical component is FADsinθF_ mADsin heta FADsinθ .
The free-body diagram for the forces acting on the watercraft is displayed in the below number.
The net force along the horizontal direction is,
FABcosθ+FADcosθ=FAFF_ mABcos heta + F_ mADcos heta = F_ mAFFABcosθ+FADcosθ=FAF
Given that, the pressure acting on the boat due to rope running A from B is equal to the pressure acting on the watercraft because of rope running A from D that is FAB=FADF_ mAB = F_ mADFAB=FAD .
mABFAB for FADF_
mADFAD in above equation.
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FABcosθ+FABcosθ=FAFFAF=2FABcosθeginarrayc\F_ mABcos heta + F_ mABcos heta = F_ mAF\\F_ mAF = 2F_ mABcos heta \endarrayFABcosθ+FABcosθ=FAFFAF=2FABcosθAns: Part A
Magnitude of pressure alengthy 3rd rope is 2FABcosθ2F_ mABcos heta 2FABcosθ .