Calculating Force Required to Drag a Carton of Books

• 8parks11
In summary: The answer should be 3.33N.In summary, the minimum force required to drag a carton of books across the floor at a constant speed, with the force applied at an angle of 45 degrees to the horizontal, is 3.33N. This takes into account the mass of the carton (40kg) and the coefficient of friction (0.60). It is important to recognize that the frictional force is proportional to the normal force, not the weight or mass of the object. This can be calculated using the equation F = μN, where μ is the coefficient of friction and N is the normal force.
8parks11
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.

This is what i tried

F=ma

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

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.

$$Fcos(\theta)=F_{friciton}=(\mu)mg$$

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.

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.

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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

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

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.

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
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

1. How do you calculate the force required to drag a carton of books?

The force required to drag a carton of books can be calculated by using the formula: force = mass x acceleration. In this case, the mass would be the weight of the carton of books and the acceleration would be the coefficient of friction between the carton and the surface it is being dragged on.

2. What is the coefficient of friction and how does it affect the force required?

The coefficient of friction is a measure of the resistance between two surfaces when they come into contact with each other. It affects the force required to drag a carton of books because a higher coefficient of friction means there is more resistance between the carton and the surface, requiring a greater force to overcome it and move the carton.

3. How do you determine the coefficient of friction for a specific surface?

The coefficient of friction for a specific surface can be determined by conducting an experiment called a friction test. This involves placing the carton of books on the surface and gradually increasing the force required to move it. The coefficient of friction can then be calculated by dividing the force required by the weight of the carton.

4. Does the weight of the carton of books affect the force required to drag it?

Yes, the weight of the carton of books does affect the force required to drag it. The heavier the carton, the greater the force required to overcome its weight and move it. This is why it is important to consider the weight of the carton when calculating the force required.

5. Are there any other factors that can affect the force required to drag a carton of books?

Yes, there are other factors that can affect the force required to drag a carton of books. These include the surface the carton is being dragged on (smooth vs rough), the shape of the carton (rectangular vs cylindrical), and any additional forces acting on the carton (such as air resistance). It is important to consider all of these factors when calculating the force required to drag a carton of books.

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