physics question....overcoming friction....?

Question

A student pulls on a rope attached to a box of books and moves the box down the hall. The studen pulls with a force of 185 N at an angle of 25.0 degrees above the horizontal. The box has a mass of 35.0 kg, and ,uk between the box and the floor is 0.27. Find the acceleration of the box.

ughgh....this is confusing me..then it says after you figure out that part.........

The student moves the box up a ramp inclined at 12 degrees with the horizontal. If the box starts from rest at the bottom of ramp and is pulled at an angle of 25.0 degrees with respect to the incline and with the same 185 N force, what is the acceleration up the ramp? assume that ,uk=0.27.

Answer 1

This is a Newton's Second Law question where you have to take into account a few different forces.

We also need to take components of forces as appropriate.

the box is not accelerating in the vertical direction, so we know the sum of forces in the vertical direction is zero. These forces are: a) the vertical component of the pull (up), b) the weight of the box (down) and c) the normal force of the floor acting on the box (up).

Since these forces balance, we have that:

N+pull(vertical) - W = 0 => N=W-pull(vertical)

the weight of the box is 35kgx9.8m/s/s=343 newtons

the vertical component of the pull is 185sin(25)=78.1 newtons,

so the Normal force is 343-78.1=264.8 N

In the horizontal direction, the forces are the horizontal component of the pull and the force of friction. In the horizontal direction:

Sum forces = Horizontal component of pull-friction = ma

the horizontal component of pull = 185cos(25)=167.7 N

friction = uk x N = 0.27x264.8 = 71.5 N

so the total force is 167.7-71.5=96.2N and this equals ma for the box, or the acceleration is:

a=96.2N/35kg = 2.75 m/s/s

Answer 2

Start with a FBD of the box
Let's look at the normal force first:

N=35*9.81-185*sin(25)
N=265.17

Now let's look parallel
185*cos(25)-N*0.27=35*a
solve for a
a=4.79 m/s^2

Additional:
Now the box is going up an incline.
First, compute the normal force:
N=35*9.81*cos(12)-185*sin(25)
N=257.66

Now, let's look parallel to the incline
185*cos(25)-N*0.27-35*9.81*sin(12)
=35*a

solve for a
a=0.763

j

Answer 3

From the second law of Newton's dynamics you obtain
Acting force in the positive direction of the plane Fcosa , friction force fmgsin in the negative direction of the plane (m mass of the body, g gravitational acceleration, f friction coefficient and so the required acceleration a is
a=(Fcosa-fmgsina)/m
As I see all the involved physical quantities are given in international units and so introduce them into the formula in order to obtain a in m/s^2.

Answer 4

Formula:
Fnet=ma

185cos25-.27(9.8)(35)=a

Answer 5

185cos(25) - (.27)(35*9.81) = 35(a)

solve for a:

a = [185cos(25) - (.27)(35*9.81)] / 35

I would solve it for you but I don't have my caclulator handy.