Expert: Ahmed Salami - 4/24/2009

Question
An SUV weighing 4600 pounds is parked on a street which has an  incline of 20 degrees. Find the force required to keep the SUV from rolling down the hill and the force of the SUV perpendicular to the hill.

Answer
Hi Brittney,
The magnitude of the forces required to keep the SUV from rolling down the hill and that perpendicular to the hill are
F(r) = mgsin#
    = 4600.sin20
    = 4600(0.3421)
    = 1573.66 pounds
and
F(p) = mgcos#
    = 4600.cos20
    = 4600(0.9397)
    = 4322.62 pounds
where mg is the weight of the SUV and # is the angle of its inclination.
Additionally, if we take our unit vectors i and j to lie on the horizontal and vertical axes. The angle of inclination of F(r) to the horizontal is 20 degrees and for F(p) is 130 degrees (where the angles are measured anti-clockwise starting from the horizontal position, a sketch would show you this because F(p) makes and angle of 70 degrees i.e 90 - 20 with the horizontal axis if it were measured clockwise starting from the left position).
In vector form,
F(r) = 1573.66(i.cos20 + j.sin20)
    = 1479i + 538j
F(p) = 4322.62(i.cos130 + j.sin130)
    = -2779i + 3312j

Note that i have only added the vector solution because i see that you'll be doing vectors which means that you can of course ignore it if you want.
Hope it helps you.

Regards