Question

A car that has a weight W is on an inclined plane with an angle of pi/12. A cable is holding the car in place ( the whole problem is shown the picture )

Answer 1

Ok, in this problem we first need to define the components parallel and perpendicular to the plane of the inclined plane, so we can project the gravitational component in each one of these. (see the image below to see the change in coordinates)

The perpendicular component will be defined by -cos(pi/12) (because the weight vector points down) and the parallel one is defined by sin(pi/12).

Now, the gravitational force will be, in magnitude, F = g*w

where g is the gravitational acceleration: g = 9.8m/s^2

Now, the perpendicular projection is:

Pe = -W*9.8m/s^2*cos(3.14/12) = W*9.799 m/s^2 (this component is canceled due to the normal force of the inclined plane)

The parallel projection will be:

Pa = W*9.8m/s^2*sin(3.14/12) = W*0.045 m/s^2

Now, we know that the car is still, then the tension on the cable must be equal in magnitude to this force (but in the opposite direction)

Then the module of the tension of the cable is:

ITI = W*0.045 m/s^2

or

T = -W*0.045 m/s^2 in the parallel axis.

b) The force that the pavement is applying on the car will be the normal force, that is equal to the perpendicular projection of the gravitational force (But again, in opposite direction).

Then the magnitude will be:

INI = W*9.799 m/s^2