Daniel O. | Math and Physics Tutor, with a math and physics degreeMath and Physics Tutor, with a math and ...

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(b) There's a vertical component of force due to the boy pulling the box. This plus the normal force exerted by the ground, F_{N}, 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 = F_{N} + 30*sin30 (where 30*sin30 is the vertical component of the force exerted by the boy)

Mykola V. | Math Tutor - Patient and ExperiencedMath Tutor - Patient and Experienced

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(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.

F_{hor}=30N*cos(30º)

F_{hor}=26N

Now that we have F_{hor} we divide it by the mass of the box to get acceleration:

a=F_{hor}/m

a=26N/28kg

a=0.93m/s^{2}

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

F_{g}=28kg*9.81m/s^{2}

F_{g}=274.9N

So since F_{N}=-F_{g} we have that F_{N}=-274.9N.

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

sinΘ=o/h

sin(30º)=F_{ver}/30N

30N*sin(30º)=F_{ver}

15N=F_{ver}

Now we just add F_{ver} and F_{N} to get the real F_{N}.

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.

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