Consider a smooth incline plane 9 m long rising to a height of 2.7 m. Calculate the largest load that can be moved up along the plane by an input force of 600 N applied parallel to the plane.
load = ___________ N
2007-12-11 05:02:27 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Physics
I get 2000N
Take 2.7/9 (which is the sine of 14.457) and multiply the by force to get the load it can move
2007-12-11 05:14:02 · answer #1 · answered by Danny-R 3 · 0⤊ 1⤋
Without friction, the only forces on the load are weight = W = mg for the load of mass m, g = 9.81 m/sec^2, and the push P = 600 N. To get the load moving, we need to accelerate it; so we have f = ma = (P - W sin(theta)) > 0 so that a > 0 and the load will move. sin(theta) = h/L; where h = 2.7 m and L = 9 m.
Then P > W sin(theta) = W (h/L) so that W < P/sin(theta) = P/(h/L) = PL/h; where everything on the RHS is known or derived. You can do the math.
You can check the logic by setting theta = 90, vertical incline, and seeing that P > W which means the push has to be a bit bigger than the weight of the load to get the load moving straight up the vertical plane. In other words, when the plane is vertical, we are simply lifting the load by overcoming its weight with the push. And that makes a lot of sense because f = ma = (P - W) > 0 so that a > 0.
BTW the ratio L/h is the so-called mechanical advantage of the wedge (the inclined plane). The wedge is one of the first and simplest machines used by mankind; the lever is the other of the first machines.
2007-12-11 05:37:44 · answer #2 · answered by oldprof 7 · 0⤊ 0⤋
The angle of inclination = 17.46 deg (I assume you know how to work it out).
The equation to find Load is:
W sin (17.46) = 600
W = 2000 N
Therefore, load = 2000 N
2007-12-11 05:09:01 · answer #3 · answered by seminewton 3 · 0⤊ 0⤋
180.0056
2007-12-11 05:05:08 · answer #4 · answered by Anonymous · 0⤊ 0⤋