# A box being pulled at an angle

by aeromat
Tags: angle, pulled
 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.
 P: 114 Mg I included as (45.0)(0.2)...?
HW Helper
Thanks
P: 10,767
 Quote by aeromat [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

 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?
 HW Helper Thanks P: 10,767 It is 205 sin(46.5) You have it already in the equation. ehild
 P: 114 Ok so the back of the book is wrong. >_< Thank you.
 P: 37 so what did you find as fn?
 P: 114 I found Fn to be 292.74N, after Subtracting appliedforceY from grav.force.
 P: 37 isnt the normal force equal to the fappy + the force of gravity though, not fappy-fg?
 P: 114 Fn + Fay - Fg = 0 so Fn = fg - Fay?
HW Helper
Thanks
P: 10,767
 Quote by aeromat I found Fn to be 292.74N, after Subtracting appliedforceY from grav.force.
It is correct. Now find the coefficient of friction.

ehild
 P: 37 dumb question, would the normal force be different if the object was pushed rather than pulled?
 HW Helper Thanks P: 10,767 Yes. The y component of the applied force would change sign. ehild

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