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# A boy pulls a box of mass 28 kg with a force of = 30 N in the direction shown in the figure below.

A boy pulls a box of mass 28 kg with a force of = 30 N in the direction shown in the figure below.

(a) Ignoring friction, what is the acceleration of the box?

(b) What is the normal force exerted on the box by the ground?

(b) There's a vertical component of force due to the boy pulling the box. This plus the normal force exerted by the ground, FN, will equal the weight (mg). If you draw a free body diagram, you'd have the weight acting down, and the normal force and the vertical component of the 30N force acting up.

mg = FN + 30*sin30      (where 30*sin30 is the vertical component of the force exerted by the boy)

FN = mg - 30*sin30

Thanks :)

11/25/2012

You're welcome :)

11/25/2012

(a) This can be found by calculating the horizontal component of the force applied and then dividing that by the mass of the box. As you may have noticed, horizontal components usually are found by using cos. it's no different here.

Fhor=30N*cos(30º)

Fhor=26N

Now that we have Fhor we divide it by the mass of the box to get acceleration:

a=Fhor/m

a=26N/28kg

a=0.93m/s2

(b) The normal force, as we recall, is equal and opposite of the force of gravity. Fg=mg; m is the mass of the box and g is the acceleration due to gravity so we have:

Fg=28kg*9.81m/s2

Fg=274.9N

So since FN=-Fg we have that FN=-274.9N.

But we also need to take into account the veritcal component of the pull on the box (this force counteracts FN). To do that we find the vertical component through the Soa part of SoaCahToa:

sinΘ=o/h

sin(30º)=Fver/30N

30N*sin(30º)=Fver

15N=Fver

Now we just add Fver and FN to get the real FN.

Enjoy! =]

Thank you, but it keeps elln me that ( b ) is wrong for some reason.

11/25/2012

Mykola, you forgot to take into account that the force from the boy pulling it has a vertical component, which reduces the normal force exerted by the ground.

11/25/2012

Ah yes, thanks for catching that, I'll fix it.

11/25/2012

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