Crate on incline?

Question

crate of mass m (=5 kg) is placed at rest on a (frictionless) inclined plane, which has an angle (= 50o) above horizontal.

a.) The force of gravity on the crate is:

size:

b.) The normal force on the crate is:

size:

c.) The net force on the crate is:

size: N

d.) How long would it take the crate to slide 6 m down the incline?

e.) If the crate were kept from sliding by a rope pulling parallel to the incline, what would the size of the tension be?

Answer 1

Hello Dorothy--

a) mass * g = 5 * 10 = 50 m/s (we're all friends so why use 9.81 m/s^2 for g, right?)  Force acts straight down perpendicular to the earth.

b)Normal force = Weight * cos 50 = mg*cos50

c) The net force on the crate is Weight * sin50 = 32.1 N acting down the plane. (The normal force of the crate on the plane is balanced by the force of the plane on the crate...although they are NOT an actio/reaction pair...but that's another question, right?)

d) The acceleration of the crate is given by a=F/m = g * cos 50 = 6.43 m/s^2.  We know dist= 6m and v0=0 so using...

t=1.37 sec (with a vf=8.78 m/sec)

You would get this from... dist = 1/2 a *t^2 +v0 t +x0 (with x0 = 0 and v0=0)

e) you must pull on the rope with a force that equals out the net force, so Tension = 32.1 N

OK, hope that helps.

-Fred