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

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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?
 
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?
 
Fn + Fay - Fg = 0
so
Fn = fg - Fay?
 
dumb question, would the normal force be different if the object was pushed rather than pulled?