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

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SUMMARY

The discussion focuses on calculating forces acting on a 45.0 kg box being pulled at an angle of 46.5° with a force of 205 N. Key calculations include determining the net horizontal force, frictional force, horizontal component of the applied force, and the coefficient of kinetic friction. The horizontal component of the applied force is confirmed to be 141.1 N, while the normal force is calculated as 292.74 N. The coefficient of friction is derived from the equation Ffr = μFn, utilizing the normal force and frictional force values.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with kinematics equations
  • Knowledge of trigonometric functions (sine and cosine)
  • Concept of friction and coefficient of kinetic friction
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  • Study the derivation of the coefficient of friction using Ffr = μFn
  • Learn about the effects of pulling versus pushing forces on normal force calculations
  • Explore advanced kinematics equations for varying acceleration scenarios
  • Investigate the impact of angle on force components in physics problems
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Students in physics courses, educators teaching mechanics, and anyone interested in understanding force calculations in applied physics scenarios.

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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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