Object dragged at an angle across a horizontal plane

[ninetyfour]

Homework Statement

Let's say I had this object that someone was dragging at an angle across a horizontal plane. I am only given the angle to the horizontal, the mass of the object, and the coefficient of kinetic friction between the object and the surface.

How do I find the applied force needed to drag the block at a constant speed?

The Attempt at a Solution

I'm not quite sure how to approach this. I feel I have too many unknowns, since the normal force also includes the y component of the applied force...

Help? Ideas?

 

[kuruman]

What is the acceleration of the object? Do you know or can you figure it out?

 

[ninetyfour]

I think the acceleration of the object is 0 because the question is asking for the applied force required to drag the object at a constant speed D:

 

[kuruman]

Correct. If the acceleration is zero, what else is also zero? Think Newton's Second Law.

 

[ninetyfour]

The net force is zero.
So the applied force equals the frictional force..
But is it that the horizontal component of the applied force equals the frictional force, or the net applied force equals the frictional force?

 

[kuruman]

The net force has two components, horizontal and vertical. Since the net force is zero, this means that the sum of all the horizontal components of the forces acting on the object is zero. The same is true for all the vertical components. So you need to write two equations

Sum of all horizontal components = 0
Sum of all vertical components = 0

This gives you two equations. How many unknowns do you have and what are they?

 

[ninetyfour]

So
Fn + Fa(y component) - mg = 0
and
Ff - Fa(x component) = 0
?

My unknowns are Fn, all components of the applied force, and the frictional force (but I do have the coefficient of kinetic friction).

 

[ninetyfour]

I know that
(I have theta)..
Fa(x) = Fa(cos)(theta)
Fa(y) = Fa(sin)(theta)
Ff = (Fn)(coefficient of kinetic friction)
Fg = 9.8m (I have m)
Fn = mg - Fa(y)

 

[kuruman]

You also have the angle. This means that the x and y components can be expressed in terms of trig functions of a known angle. In other words, only the magnitude of the pulling force is unknown. As far as kinetic friction is concerned, is there an expression that gives you what it is in terms of the coefficient of kinetic friction?

 

[ninetyfour]

Ff = (Fn)(coefficient of kinetic friction)

I think my problem is putting it all together :S
Some of my equations will have too many unknowns...sometimes I have one unknown but it cancels out and the equation becomes useless...

 

[kuruman]

Can you put it together? What are the equations and what are the unknowns? There should be two and two.

 

[kuruman]

So
Fn + Fa(y component) - mg = 0
and
Ff - Fa(x component) = 0

Work on the second equation first. Put in the angle and the expression for the force of friction.

 

[ninetyfour]

Okay,
So I tried again.

Here are the knowns for my question:
m = 21kg
theta = 40 degrees to the horizontal
coefficient of kinetic friction = 0.2
- constant speed
Applied force = ?

After a bunch of work...
Fa = 46 N

Is this correct?? That to move this block at a constant speed with those givens, the applied force will be 46 N?

 

[kuruman]

That is correct, I checked the numbers.

 

[ninetyfour]

THAANNKK YOUUUU !
I really appreciate your help :)