Calculating Frictional Force and Time to Move a Box Across a Floor

In summary, the box moves 5 meters in 1.58 seconds when the force is 445 N (roughly the weight of a pre-teen girl), but it takes almost 16 m/s^{2} (30 mph/s) to move the box with the full force of 455 N.
  • #1
Fanjoni
11
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A box of books weighing 229 N is shoved across the floor by a force of 455 N exerted downward at an angle of 35° below the horizontal.

(a) If µk between the box and the floor is 0.57, how long does it take to move the box 5 m, starting from rest?

(b) If µk between the box and the floor is 0.75, how long does it take to move the box 5 m, starting from rest?

I started by finding the x and y vectors x=272.71 y=260.98 Then i tried to th Neutral force and used it in the equation Fk=µk*N But i am lost and don't know hwat to do please help me
Thanks
 
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  • #2
Fanjoni said:
A box of books weighing 229 N is shoved across the floor by a force of 455 N exerted downward at an angle of 35° below the horizontal.

(a) If µk between the box and the floor is 0.57, how long does it take to move the box 5 m, starting from rest?

(b) If µk between the box and the floor is 0.75, how long does it take to move the box 5 m, starting from rest?

I started by finding the x and y vectors x=272.71 y=260.98 Then i tried to th Neutral force and used it in the equation Fk=µk*N But i am lost and don't know hwat to do please help me
Thanks

Draw a free body diagram. Remember that N is not simply mg, but also the magnitude of the y-component of the acting force.

[tex]F_{k}=\mu_{k}N[/tex]
[tex]|N|=mg+F_{push}\sin(\theta)[/tex]

You know mg (229 N), [itex]F_{push}[/itex] (455 N), and [itex]\theta[/itex] (35[itex]^{\circ}[/itex] SE).
 
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  • #3
i found acceleration 15.74 and then found time to be 1.694s is this correct?
 
  • #4
Fanjoni said:
i found acceleration 15.74 and then found time to be 1.694s is this correct?

I haven't solved the problem, so I don't know if that answer is correct. However, you should ask yourself, does an acceleration of almost 16 [itex]m/s^{2}[/itex] seem reasonable? That's over 30 mph/s, and the box is moving 5 meters in under 2 seconds. I'd say it's probably not correct.

Show your work, and we can pick out the error.
 
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  • #5
geoffjb said:
I haven't solved the problem, so I don't know if that answer is correct. However, you should ask yourself, does an acceleration of almost 16 [itex]m/s^{2}[/itex] seem reasonable? That's over 30 mph, and the box is moving 5 meters in under 2 seconds. I'd say it's probably not correct.

Show your work, and we can pick out the error.

Actually, it sounds reasonable. Look at the forces involved on a not so massive item.

EDIT: cos(35) * 455N * 9.8m/s^2 / 229N = 15.95
 
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  • #6
moose said:
Actually, it sounds reasonable. Look at the forces involved on a not so massive item.

EDIT: cos(35) * 455N * 9.8m/s^2 / 229N = 15.95

You're saying it's reasonable for a force of 445 N (roughly equal in magnitude to the weight of a pre-teen girl)[itex]-[/itex]and not the full force, but a component of it[itex]-[/itex]to accelerate a 50 pound object to 60 mph in 2 seconds?
 
  • #7
geoffjb said:
You're saying it's reasonable for a force of 445 N (roughly equal in magnitude to the weight of a pre-teen girl)[itex]-[/itex]and not the full force, but a component of it[itex]-[/itex]to accelerate a 50 pound object to 60 mph in 2 seconds?
Yes.
If it were the full force, 455N, it would accelerate the 23.4kg object at 19.5 m/s^2.

Think about it. The acceleration due to gravity is 9.8m/s^2, correct? That's nearly 22 mph per second. Now, a 229N object in would experience 229N force from gravity, by definition. Soooo, a greater force (hence, a greater than half portion of 455N), would exert a greater acceleration. That's the intuition part that you're arguing with me about. The math works out perfectly as well ;)EDIT: That acceleration is without any friction as well. Which is what I believe he was originally aiming at?
 
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  • #8
moose said:
Yes.
If it were the full force, 455N, it would accelerate the 23.4kg object at 19.5 m/s^2.

Think about it. The acceleration due to gravity is 9.8m/s^2, correct? That's nearly 22 mph per second. Now, a 229N object in would experience 229N force from gravity, by definition. Soooo, a greater force (hence, a greater than half portion of 455N), would exert a greater acceleration. That's the intuition part that you're arguing with me about. The math works out perfectly as well ;)


EDIT: That acceleration is without any friction as well. Which is what I believe he was originally aiming at?

Yes, you're absolutely right.

As for the original poster, I calculated an acceleration of 4m/s/s, not 16. The time is close to my value.
 
  • #9
The frictional force opposing movement is 0.57*(229 + 261)N
=279.3 N

ma = 372.7 - 279.3 = 93.4 Newtons,the mass is 229/9.8 = 23.4 Kg

a=93.4N/23.4Kg = 4 meters/sec^2

Since at^2/2 = distance ----------> sqrt(2*distance/a) = t = 1.58 sec


yes yes i finaly got it thanks guys
 

FAQ: Calculating Frictional Force and Time to Move a Box Across a Floor

1. What does the weight of the box mean?

The weight of the box refers to the force exerted by the Earth's gravity on the box. It is measured in newtons (N) and is a measure of the box's mass and the strength of the gravitational field it is in.

2. How is the weight of the box calculated?

The weight of the box is calculated by multiplying the mass of the box (in kilograms) by the acceleration due to gravity (9.8 m/s²). This results in the weight in newtons (N).

3. What does the angle (35°) represent in this scenario?

The angle (35°) represents the angle at which the box is being held or placed. It is an important factor in calculating the weight as it affects the vertical and horizontal components of the force being exerted on the box.

4. Can the weight of the box change at different angles?

Yes, the weight of the box can change at different angles. This is because the angle affects the direction and magnitude of the force being exerted by gravity on the box. The weight will be greatest when the box is held vertically (at 90°) and will decrease as the angle decreases.

5. How does the weight of the box at 35° compare to its weight at 90°?

The weight of the box at 35° will be less than its weight at 90°. This is because at 35°, the weight is being split into two components - a vertical component (which is equal to the weight at 90°) and a horizontal component. The horizontal component reduces the overall weight of the box at 35°.

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