Force PROBLEM

What minimum force is required to drag a carton of books across the floor at consant speed if the force is applied at an angle of 45 degrees to the horizontal? Take the mass of the carton 40 kg and the coefficient friction as 0.60.


Someone PLEASE help me..

This is what i tried

F=ma

now
F=40xmxsin45
so i did 40m1/2
which F=20m
and it got really confusing

can someone give me a solution and answer please
 
You have to take friction into account, which you didn't. And its cos45 not sin for the record, but it doen't really matter since they are the same.

Moreover sin45 = 1/Surd2 not 1/2
 
thanks for the help so here is this right?

F=40(.60)cos45???
 
You're close. Friction is proportional to the WEIGHT, not the mass. You are using mass there.
 
[tex]Fcos(\theta)=F_{friciton}=(\mu)mg[/tex]
 
F x cos45= 9.8(40).60

correct? so the answer would be 3.326?
 
That equation is correct. If your maths is correct then your answer is also correct. Just remeber to draw a free body diagram if you get stuck.
 

OlderDan

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8parks11 said:
F x cos45= 9.8(40).60

correct? so the answer would be 3.326?
This is not correct. The frictional force is NOT proportional to weight. It is proportional to the normal force and in this problem the normal foce is not the weight of the carton.

Look again at the forces acting in the vertical direction. There is gravity (weight), the normal force, and one more.
 
Last edited:
That is not correct. The weight force is NOT normal force and is therefore not proportional to the friction. You'll have to take into consideration on the tension force as well. The key point here is to recognize the net force for the y direction. It's pretty recognizable...
 
cy, the answer is correct. reread the problem if u want to know why ur explanation is invalid
 

OlderDan

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8parks11 said:
cy, the answer is correct. reread the problem if u want to know why ur explanation is invalid
It is not correct. Your computation using your equation is off by a factor of 100, and even if that is fixed it is the wrong answer. If you changed the angle from 45 degrees to 60 degrees, your equation would give a force that is greater than the weight of the object and enough so that even applied at that angle you would be lifting the carton off the floor.
 
OlderDan and cy are right. The friction is not proportional to the weight in this problem. You have to take into consideration that F has a vertical component. It´s pushing the object into the ground so friction is added there...
 
just to clear this up, there is actually a figure with a FBD


it shows Ffr,mg, F cos 45, Fsin45, F

The question:
What minimum force is required to drag a carton of books across the floor at consant speed if the force is applied at an angle of 45 degrees to the <<<<<<horizontal>>>>>>>? Take the mass of the carton 40 kg and the coefficient friction as 0.60.



F cos 45 = 40 (.60)g
Fcos 45 - Ff
a=0
Fcos45=Ff=muN
F=Mumg/cos45
0.60(40)(9.8) / cos 45
isnt this the answer?
 

OlderDan

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student85 said:
OlderDan and cy are right. The friction is not proportional to the weight in this problem. You have to take into consideration that F has a vertical component. It´s pushing the object into the ground so friction is added there...
The carton is being dragged, not pushed. The force is upward at 45 degrees.
 

OlderDan

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8parks11 said:
just to clear this up, there is actually a figure with a FBD


it shows Ffr,mg, F cos 45, Fsin45, F

The question:
What minimum force is required to drag a carton of books across the floor at consant speed if the force is applied at an angle of 45 degrees to the <<<<<<horizontal>>>>>>>? Take the mass of the carton 40 kg and the coefficient friction as 0.60.



F cos 45 = 40 (.60)g
Fcos 45 - Ff
a=0
Fcos45=Ff=muN
F=Mumg/cos45 <== This is not correct. N is not mg
0.60(40)(9.8) / cos 45
isnt this the answer?
N is not mg. Look at the vertical forces.
 
Last edited:
8parks11 said:
cy, the answer is correct. reread the problem if u want to know why ur explanation is invalid
No. As I said, friction is proportional to weight (some people like the term apparent weight for this type of situation). It is not proportional to the mass. It is not proportional to mg. But everyone agrees that it is proportional to the normal force, which is not mg, and that is the point.

Dorothy
 

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