How Do You Calculate Forces on a Box Pulled at an Angle?

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Homework Help Overview

The discussion revolves around calculating forces acting on a box being pulled at an angle, specifically addressing the net force, frictional force, horizontal component of the applied force, and the coefficient of kinetic friction. The problem involves concepts from dynamics and kinematics.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the calculation of acceleration and the summation of forces in both horizontal and vertical directions. There are attempts to clarify the role of the applied force's components in the force equations.

Discussion Status

Some participants have provided calculations and interpretations regarding the forces involved, while others question the assumptions made about the normal force and its relationship to the applied force. There is an ongoing exploration of the coefficient of friction based on the derived normal force.

Contextual Notes

Participants discuss potential discrepancies with textbook answers and the implications of pulling versus pushing the box on the normal force. The conversation reflects uncertainty about the correct application of force components in the equations.

aeromat
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Homework Statement


A 45.0 kg box is pulled with a force of 205 N
by a rope held at an angle of 46.5° to the
horizontal. The velocity of the box increases
from 1.00 m/s to 1.50 m/s in 2.50 s. Calculate
a) Net force acting horizontally on the box
b) Frictional force on the box
c) Horizontal component of the applied force
d) Co-efficient of kinetic friction between the box and the floor


Homework Equations


Kinematics equations
[mu][Fn] = frictional force


The Attempt at a Solution


pg55-ChapRev1-29.png

a) I found acceleration by: change in velocity = 0.5m/s, change in time = 2.50s
v
---
t
= 0.2m/s^2

Ok I found the summation of the X and Y forces:
b) Summation of X forces: [(205)(cos46.5) - frictionalF = (45)(0.2)]
frictionalF = 132.1N [<---]

Summation of Y forces: [Fn + (205)(sin46.5) - (45.0)(0.2) = 0]
normalF = 292.74N

c) Now, I found the horizontal component of the applied force to be:
[205][cos46.5] = 141.1N, but the answer at the back says 293N. Have I done something wrong?

d) I am not sure how I am going to get the coefficient.
 
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Mg I included as (45.0)(0.2)...?
 
aeromat said:


Summation of Y forces: [Fn + (205)(sin46.5) - (45.0)(0.2) = 0]

The horizontal component of the applied force is 205 cos(46.5)= 141.1 N it is correct.

The coefficient of friction is obtained from
Ffr=μ Fn. Calculate Fn from the equation for the y components of forces.

ehild
 
Last edited:
Doesn't the applied force y-component count as one of the forces in the Y-summation?
 
It is 205 sin(46.5) You have it already in the equation. ehild
 
Ok so the back of the book is wrong. >_< Thank you.
 
so what did you find as fn?
 
I found Fn to be 292.74N, after Subtracting appliedforceY from grav.force.
 
isnt the normal force equal to the fappy + the force of gravity though, not fappy-fg?
 
  • #10
Fn + Fay - Fg = 0
so
Fn = fg - Fay?
 
  • #11
aeromat said:
I found Fn to be 292.74N, after Subtracting appliedforceY from grav.force.

It is correct. Now find the coefficient of friction.

ehild
 
  • #12
dumb question, would the normal force be different if the object was pushed rather than pulled?
 
  • #13
Yes. The y component of the applied force would change sign.

ehild
 

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