Person dragging a bow with a rope - Calculate friction

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SUMMARY

The discussion focuses on calculating the force exerted by a person dragging a box with a mass of 60 kg, initially moving at 4 m/s and accelerating to 10 m/s over 3 seconds. The coefficient of friction between the box and the ground is 0.7. Participants clarify that the frictional force is calculated as the product of the coefficient of friction and the normal force, which equals the weight of the box. The correct approach involves using Newton's second law, F = ma, to find the net force and considering friction's impact on the pulling force.

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  • Understanding of Newton's laws of motion
  • Knowledge of friction and coefficients of friction
  • Ability to calculate acceleration from velocity and time
  • Familiarity with basic physics equations, such as F = ma
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  • Study the concept of normal force and its relation to weight
  • Learn how to calculate frictional force using the coefficient of friction
  • Explore kinematic equations for motion analysis
  • Investigate the implications of net force in dynamic systems
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Students studying physics, particularly those focusing on mechanics, as well as educators seeking to clarify concepts related to force, friction, and motion.

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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, \mu is the constant relating the frictional force between two surfaces to the normal contact force between the two surfaces.
F_{friction} = \mu N​

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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