A block of mass m = 2.0 kg is held in equilibrium on an incline of angle = 60° by the horizontal force F (ignore friction)
Determine the value of F
Determine the normal force exerted by the incline on the block.
the diagram looks like a triangle with a box on top of the triangle with F exerted on the box.
how do u solve this?
2006-10-17 16:51:23 · 3 answers · asked by tingerpoo 2 in Science & Mathematics ➔ Physics
W = m g
W = 2 kg (9.8 m/sec^2)
W = 19.6 N, straight down
W* = W sin(60)
W* = 16.974 N, slope down
Therefore...
F* = 16.974 N, slope up
F = F* / cos(60)
F = 33.948 N, horizontally
I'll combine the equations, so you can do it faster next time...
F = mg tan(incline angle)
If you think about it, this equation MAKES SENSE. Why? The steeper the slope of a frictionless surface is, the more horizontal force you're going to need to hold a mass on it.
When the slope is 90 degrees (i.e., a vertical wall), you'd need an infinite horizontal force to keep it there. What's the tangent of 90 degrees? Infinity.
When the slope is zero (i.e., a table top), the block doesn't have a down direction to slide toward, so the required force is also zero. What's the tangent of zero? Zero.
Always look for how a physics problem makes sense. That's how you tell a good proposed answer from a pile of horse crap from some freshman who just wants you to think that he's smart too.
2006-10-17 19:30:22 · answer #1 · answered by David S 5 · 0⤊ 0⤋
(2kg)cos(60) is F
the normal force is just the force that the incline is exerting on the box so i believe that would be:
(2kg)sin(60)
2006-10-18 00:03:58 · answer #2 · answered by Elfy 2 · 0⤊ 0⤋
Think trigonometry.
2006-10-17 23:58:26 · answer #3 · answered by Brian L 7 · 0⤊ 0⤋