Question

The student in item 1 moves a box up a ramp inclined at 12 degrees with the horizontal. If the box starts from rest at the bottom of the 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?

Answer 1

The box is on the plane inclined at angle of 12 degree. The component of the weight of the box perpendicular to the incline is mg cos 12. = 335.5 N. The normal reaction force acting on the box is 335.5- 78.2 = 257.3 N.

Answer 2

Given:

The angle of inclination of the ramp with respect to horizontal = 12°

The angle at which the box is pulled with respect to the inclined = 25°

The force with which box is pulled = 185 N

To find:

Acceleration of the box

Solution:

The force applied in pulling (F) = 185 N

The angle at which the force is applied = 25°

We will break this force into horizontal and vertical components

The horizontal component of force parallel to incline = 185 cos(25)

Fx = 167.66 N

The vertical component of force perpendicular to incline = 185 sin(25)

Fy = 78.184 N

Now let the mass of box be m

Force due to weight of box = mg N

This force makes an angle of 12° with the perpendicular to the incline.

Now we will break this force into horizontal and vertical components.

The horizontal component of force parallel to incline = mg cos(12)

mgx = 0.97mg N

The vertical component of force perpendicular to incline = mg sin(12)

mgy = 0.20mg N

Now,

total force in horizontal direction = Fx - mgy

total force in vertical direction = mgy - fy

Therefore acceleration will be

acceleration =  force/mass

=  $\frac{Fx - mgy}{mgy - Fy}$

= $\frac{167.66 - 0.97mg}{0.20mg - 78.18}$

Putting the value of m we will get the values of acceleration.