A box weighing 229 N at 35°

Fanjoni

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 dont know hwat to do please help me
Thanks

geoffjb

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).

Fanjoni

i found acceleration 15.74 and then found time to be 1.694s is this correct?

geoffjb

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.

moose

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

geoffjb

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?

moose

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?

geoffjb

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.

Fanjoni

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