A box being pulled at an angle

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In summary: The coefficient of friction between the box and the floor is .In summary, the force applied to a 45.0 kg box is 205 N and it increases from 1.00 m/s to 1.50 m/s in 2.50 s. The horizontal component of the applied force is 141.1 N.
  • #1
aeromat
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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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  • #2
Mg I included as (45.0)(0.2)...?
 
  • #3
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:
  • #4
Doesn't the applied force y-component count as one of the forces in the Y-summation?
 
  • #5
It is 205 sin(46.5) You have it already in the equation. ehild
 
  • #6
Ok so the back of the book is wrong. >_< Thank you.
 
  • #7
so what did you find as fn?
 
  • #8
I found Fn to be 292.74N, after Subtracting appliedforceY from grav.force.
 
  • #9
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
 

What is the concept of "A box being pulled at an angle"?

The concept of "A box being pulled at an angle" refers to the situation where a box or object is being pulled or moved in a direction that is not directly in front of it, but at an angle.

How does the angle affect the force required to pull the box?

The angle at which the box is being pulled affects the amount of force required to move it. The force can be broken down into two components - the horizontal force and the vertical force. The steeper the angle, the greater the vertical force and therefore, the greater the force needed to move the box.

What is the relationship between the angle and the friction of the box?

The angle also affects the friction of the box. As the angle increases, the surface area of contact between the box and the surface decreases, resulting in less friction. This means that the box will slide more easily at steeper angles compared to shallow angles.

How does the weight of the box impact its movement at an angle?

The weight of the box plays a role in its movement at an angle. The weight, or mass, of the box affects the amount of horizontal force required to move it. A heavier box will require more force to move at the same angle compared to a lighter box.

What factors should be considered when pulling a box at an angle?

When pulling a box at an angle, several factors should be considered, such as the weight and friction of the box, the angle at which it is being pulled, and the force needed to move it. It is important to assess these factors in order to successfully move the box without causing any unnecessary strain or injury.

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