Person dragging a bow with a rope - Calculate friction

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Homework Help Overview

The problem involves a person dragging a box with a mass of 60 kg using a rope, with a focus on calculating the force exerted while considering friction. The scenario describes the box's initial and final velocities over a time period, along with the coefficient of friction between the box and the ground.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the definition of the coefficient of friction and its role in calculating the frictional force. There are attempts to relate known quantities such as mass, initial and final velocities, and time to find the applied force. Questions arise about the correct equations to use and the distinction between mass and weight.

Discussion Status

The discussion is ongoing, with participants exploring various interpretations of the problem. Some have provided guidance on using acceleration and net force, while others are clarifying concepts related to friction and normal force. There is no explicit consensus yet on the approach to take.

Contextual Notes

Participants note confusion regarding the distinction between mass and weight, as well as the correct application of the coefficient of friction in the context of the problem.

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Homework Statement


The question says
"A person is dragging a box with the mass of 60kg by a rope. The person begins to drag the box at 4 m/s [F]. All of a sudden the person sees on coming traffic. The person calculates that they have 3 seconds to get out of the way. If at the end of the 3 seconds the box is moving 10 m/s [F] and the coefficient of friction between the box and the ground is 0.7, find the force which the person is pulling the box."

my biggest problem is I don't know what a coefficient is.

Homework Equations


im not to sure what equations i would need to use. My guess would be d=(vi)(t) + 1/2 (a)(t^2)


The Attempt at a Solution


To find out the force which the person is pulling the box would we still use
F= m*a

and we would need to figure out the displacement first with d=(vi)(t) + 1/2 (a)(t^2)
 
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The coefficient of friction, [tex]\mu[/tex] is the constant relating the frictional force between two surfaces to the normal contact force between the two surfaces.
[tex]F_{friction} = \mu N[/tex]​

The velocity of the box increases from 4m/s to 10m/s in 3s. I presume that should be enough for you to obtain the average force exerted on the box during this time period using the standard F=ma?
 


so μ is 0.7?
 


yep.
 


How would we begin this question
my known's are
mass= 60kg
vi=4 m/s [F]
v2=10m/s [F]
t= 10s
friction=0.7

but i don't know what equation to use to find the applied force
 


you have time, v0 and vf. from there, you must solve for acceleration, then plug it into F = ma , which will be your net force. From there, you must consider all the different forces and single out the pulling force. (Hint: the pulling force is not all that is affecting the pull.. what does friction do?)
 


dois said:
How would we begin this question
my known's are
mass= 60kg
vi=4 m/s [F]
v2=10m/s [F]
t= 10s
friction=0.7

but i don't know what equation to use to find the applied force


friction is not 0.7, that's the coefficient of friction. Friction is the coefficient of friction multiplied by the normal force
 


ok, and the normal force is equal to the mass right?
so then it would be
0.7 x 60
which would then be
42?
 


dois said:
ok, and the normal force is equal to the mass right?
so then it would be
0.7 x 60
which would then be
42?

the normal force is what's keeping the object "up", it is the sum of vertical components.. so in this example, you only have the force of gravity or the weight pushing it down to the ground, so the normal force is equal to the weight
 
  • #10


Alright. So then the normal force is 60, because the weight is 60kg.
and then friction would be, 0.7x60=42
is that right?
 
  • #11


dois said:
Alright. So then the normal force is 60, because the weight is 60kg.
and then friction would be, 0.7x60=42
is that right?

no, mass and weight are two different things. I think that you should revise your concepts
 

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