Find the force required to keep a 17,000 pound truck parked on a hll that makes an angle of 4 degrees, 20 minutes with the horizontal.
Not sure I did it correctly! Solution and steps much appreciated. Thanks!
2006-11-18 02:45:01 · 3 answers · asked by mdetaos 3 in Science & Mathematics ➔ Mathematics
Just like any other weight on an incline problem, draw a right triangle. However, the lines aren't distances, they're force vectors. The vertical vector = 17,000. The horizontal line is irrelevant. The hypotenuse is the force vector you want. It represents the force of the truck sliding down the hill. Basically, find this vector, and that's how much force of friction it takes to prevent sliding (so net force on the truck = 0).
The angle between the horizontal line of the triagle and the hypotenuse is 4 degrees, 20 minutes. You want the other (non-right) angle. SInce all three angles add up to 180 degrees, 180 - 4degs/20min (change that to decimals to save headaches) = the angle you need.
Now, cos(that angle) = 17,000 / (the value of the vector your solving for, all it X). Rearranging that gives you X = 17000/cos(that angle).
Happy solving.
2006-11-18 03:02:15 · answer #1 · answered by marty.smith01 2 · 1⤊ 0⤋
Dupinder jeet kaur k's solution is correct, BUT she inserted a factor of 9.8 in the equation. You should delete that (or replace it with a 1).
Reason: 9.8 is the factor to convert to Newtons when the mass is given in kilograms. In this case, you gave the weight of the truck, which is already a force, so no conversion from mass to force is needed.
2006-11-18 11:42:55 · answer #2 · answered by actuator 5 · 0⤊ 0⤋
it is equal to component of
weight along the slope
=mg*sin(angle of inclination)
=(17000/0.4536)*9.8*sin(4deg20min)
2006-11-18 10:55:51 · answer #3 · answered by Dupinder jeet kaur k 2 · 0⤊ 0⤋