Physics problem that i cannot figure out?

I cannot figure out how to solve this problem...


A machine in an ice factory is capable of exerting 3.00x10^2 N of force to pull a large block of ice up a slope. The block weighs 1.22 x 10^4 N. Assuming there is no friction, what is the maximum angle that the slop can make with the horizontal if the machine is able to complete the task?Physics problem that i cannot figure out?
Gravity always works directly downward, so when an item is traveling on a slope, the gravity will present itself as two components: parallel and perpendicular to the slope. The perpendicular component is usually offset by the normal force (in this case it is). The machine is pulling opposite the parallel component of the force of gravity. So, in order to find maximum angle, you set the maximum force of the machine equal to the parallel component of gravity. If you diagram it, you will find that the parallel component is equal to mgsinz (i'm calling the angle z, because i can't type a theta).


that all leads me to:


3.00x10^2 N = 1.22 x 10^4 N * sinz


sinz = 0.0246


z = 1.41 degrees.





Sorry about the wordy answer, it would have been a lot easier if i could have diagrammed it.Physics problem that i cannot figure out?
The first thing you always have to do with this type of problem is draw your free body diagram. split the gravitational force on the block into components perpendicular to and parallel to the slope. (you can draw a right angled triangle on the gravity vector with the straight down gravity vector as the hypoteneuse) the component that is parallel to the slope should be equal to the force of the machine on the block. that is, 3.00*10^2 N = (1.22*10^4N)(sin(theta)). and then just solve for theta.
the component of the weight parallel the ramp is 1.22x10^4(sin(theta))


if the machine is pulling the block in a manner completely parallel the ramp, then:


F = ma


F(machine,up the ramp) - F(weight, down the ramp) = ma


300N - 12200Nsin(theta) = ma





if the machine is going to move the block up the ramp, then a %26gt; 0 so


ma %26gt; 0


300N - 12200Nsin(theta) %26gt; 0


300N %26gt; 12200Nsin(theta)


300/12200 %26gt; sin(theta)


theta %26lt; sin-1(3/122)