force parallel to the incline and down the hill = 600sin(30) = 300 lbs
A 600 lb. wheel is set on a ramp inclined 30 degrees. What is the force required to keep the wheel from rolling down the ramp?
I set the problem like this:
And I thought the side labeled 'x' is the force needed to keep the wheel up there, which, if calculated by sin(3) = 600/x is 1,200 lbs. but that's not the correct solution.
What am I doing wrong here?
How is it understood that the object is being pulled into the triangle? I thought gravity was pulling on the object straight downwards, and the weight of the object was the force being slid down the ramp, so wouldn't it be the combined horizontal and vertical forces that's making the object go down the ramp?
the component of weight parallel to the incline is the force that pulls the object down the ramp ... the perpendicular component of weight is balanced by the force of the incline's surface (also known as the normal force) pushing up perpendicular to the incline.