A 50 N crate is pulled up a 5 m inclined plane by a worker at constant velocity. If the plane is inclined at an angle of 37 degree and there exists a constant force of 10 N between the crate and the surface, what is the force applied by the worker?
I think you would go:
F= -10 +50N-9.8 to get 30 N
am I right? please help! thanks
2007-02-13 11:16:42 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
The force is:
50*sin(37)+10
=40 N
As long as the force is parallel to the plane.
j
2007-02-13 11:35:05 · answer #1 · answered by odu83 7 · 0⤊ 0⤋
Since the crater moves with constant velocity,
the net force acting on the crater must be zero.
The component of the gravitational force that is pulling down the crater along the plane is 50 sin 37 = 30.09 N.
Assuming that the 10 N force is also pulling the crate down along the plane
the total downward force is 10 + 30 .09 along the plane
The man must apply a force of 40.09 N up along the plane to make the crate move with constant velocity.
2007-02-14 12:02:37 · answer #2 · answered by Pearlsawme 7 · 0⤊ 0⤋
I am afraid you are wrong.
I am uncomforablt with the 50N crate thing, are you sure it isnt 50 kg?
In theory, here are the forces:
10N friction
force of gravity is mass*accel due to gravity * sin(38) so: 50kg*9.81m/S^2*sin(38)=302N
So the guy has to pull 302N+10N=312N
2007-02-13 19:41:38 · answer #3 · answered by Anonymous · 0⤊ 0⤋