A box being pulled at an angle


by aeromat
Tags: angle, pulled
aeromat
aeromat is offline
#1
Feb26-11, 10:06 AM
P: 114
1. The problem statement, all variables and given/known data
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


2. Relevant equations
Kinematics equations
[mu][Fn] = frictional force


3. The attempt at a solution

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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aeromat
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#2
Feb26-11, 10:23 AM
P: 114
Mg I included as (45.0)(0.2)...?
ehild
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#3
Feb26-11, 10:27 AM
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Quote Quote by aeromat View Post
[b]

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

aeromat
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#4
Feb26-11, 10:32 AM
P: 114

A box being pulled at an angle


Doesn't the applied force y-component count as one of the forces in the Y-summation?
ehild
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#5
Feb26-11, 10:35 AM
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It is 205 sin(46.5) You have it already in the equation.


ehild
aeromat
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#6
Feb26-11, 10:43 AM
P: 114
Ok so the back of the book is wrong. >_< Thank you.
PhysicsAdvice
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#7
Feb26-11, 10:47 AM
P: 37
so what did you find as fn?
aeromat
aeromat is offline
#8
Feb26-11, 10:50 AM
P: 114
I found Fn to be 292.74N, after Subtracting appliedforceY from grav.force.
PhysicsAdvice
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#9
Feb26-11, 10:54 AM
P: 37
isnt the normal force equal to the fappy + the force of gravity though, not fappy-fg?
aeromat
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#10
Feb26-11, 11:14 AM
P: 114
Fn + Fay - Fg = 0
so
Fn = fg - Fay?
ehild
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#11
Feb26-11, 02:25 PM
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Quote Quote by aeromat View Post
I found Fn to be 292.74N, after Subtracting appliedforceY from grav.force.
It is correct. Now find the coefficient of friction.

ehild
PhysicsAdvice
PhysicsAdvice is offline
#12
Feb26-11, 10:10 PM
P: 37
dumb question, would the normal force be different if the object was pushed rather than pulled?
ehild
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#13
Feb26-11, 10:14 PM
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P: 9,834
Yes. The y component of the applied force would change sign.

ehild


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