Register to reply 
A box being pulled at an angle 
Share this thread: 
#1
Feb2611, 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) Coefficient 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. 


#2
Feb2611, 10:23 AM

P: 114

Mg I included as (45.0)(0.2)...?



#3
Feb2611, 10:27 AM

HW Helper
Thanks
P: 10,384

The coefficient of friction is obtained from F_{fr}=μ F_{n}. Calculate Fn from the equation for the y components of forces. ehild 


#4
Feb2611, 10:32 AM

P: 114

A box being pulled at an angle
Doesn't the applied force ycomponent count as one of the forces in the Ysummation?



#5
Feb2611, 10:35 AM

HW Helper
Thanks
P: 10,384

It is 205 sin(46.5) You have it already in the equation.
ehild 


#6
Feb2611, 10:43 AM

P: 114

Ok so the back of the book is wrong. >_< Thank you.



#7
Feb2611, 10:47 AM

P: 37

so what did you find as fn?



#8
Feb2611, 10:50 AM

P: 114

I found Fn to be 292.74N, after Subtracting appliedforceY from grav.force.



#9
Feb2611, 10:54 AM

P: 37

isnt the normal force equal to the fappy + the force of gravity though, not fappyfg?



#10
Feb2611, 11:14 AM

P: 114

Fn + Fay  Fg = 0
so Fn = fg  Fay? 


#11
Feb2611, 02:25 PM

HW Helper
Thanks
P: 10,384

ehild 


#12
Feb2611, 10:10 PM

P: 37

dumb question, would the normal force be different if the object was pushed rather than pulled?



#13
Feb2611, 10:14 PM

HW Helper
Thanks
P: 10,384

Yes. The y component of the applied force would change sign.
ehild 


Register to reply 
Related Discussions  
Friction, object pulled at an angle, given only (mue) and max tension... hmmm...  Introductory Physics Homework  5  
Angle of incidence, angle of refraction and angle of reflection  Introductory Physics Homework  1  
Pulled box  Introductory Physics Homework  3  
Block pulled at an angle with friction  Introductory Physics Homework  4  
Force of water on a pulled barge with angle  Introductory Physics Homework  2 