A block of mass m = 2 kg is held in equilibrium on an incline of angle of 60 degrees by a horizontal force.
a. Determine the magnitude of this horizontal force.
b. Determine the magnitude of the normal force on the block.
2006-12-03 20:23:56 · 5 answers · asked by Some Guy 2 in Science & Mathematics ➔ Physics
Draw a force diagram and get busy ☺
The gravitational force (straight down) is simply the mass times the acceleration of gravity (F=ma ☺) so
2*9.8 = 19.6N. But the mass is supported by a plane inclined at 60° to the horizontal and so this force can be broken up into two orthogonal components. One of them is normal to the plane and the other is parallel to the plane. The normal force is F_n = F*cos(60°) so
F_n = 19.6*.5 = 9.8N. The parallel force is
F_p = F*sin(60°) = 19.6*.866 = 16.97N.
Now, a force applied horizontally to the mass must produce 16.97N parallel to the plane to keep the mass in equilibrium and a horizontal force will have a parallel component of F*cos(60°) so
16.97 = F*.5 => F = 33.94N. This is the horizontal force needed to keep the mass in equilibrium. The normal force created by this is F*sin(60°) = 33.94*.866 which is
29.39N normal to the plane. The total normal force is the sum of the two normal forces so
9.8 + 29.39 = 39.19N normal to the plane.
That wasn't so bad, was it ? ☺
Doug
2006-12-03 20:58:47 · answer #1 · answered by doug_donaghue 7 · 0⤊ 0⤋
The angle of the inclined plane is 60 degree.
A normal to this plane will also tilt through 60 degree from the vertical.
Resolving the normal reaction R (which is normal to the plane),
The vertical component is R cos 60 (since it is adjacent to the angle 60 degree)
The horizontal component is R sin 60.
For equilibrium
R sin 60 = F (the horizontal force)
R cos 60 = = mg (the weight)
Dividing
Tan 60 = F /mg.
F = mg tan 60.
= 2 x 9.8 x tan 60 = 34 N
R = F/ sin 60 = 39.25 N.
2006-12-03 22:19:45 · answer #2 · answered by Pearlsawme 7 · 0⤊ 0⤋
Do you really want an answer to this question or simply want to test the ability of your answerers? This is elementary Mechanics which every first grader in Maths should know about. Anyway, here is the answer:
The normal force on the block is given by the eqn
Ncos60 = 2kg , N= 2/cos60 = 4 kg
The horizontal force H is given by the eqn Nsin60 = H, from where H = 3.46 kg
2006-12-03 21:16:02 · answer #3 · answered by Paleologus 3 · 0⤊ 0⤋
A "centrifugal rigidity" might want to be a rigidity that acts immediately remote from the middle. There are some information, yet actual there is not any such element as centrifugal rigidity. What maximum persons imagine of as centrifugal rigidity is very an artifact of inertia. A "centripetal rigidity" is a rigidity that acts in the direction of the middle. Any merchandise that is transferring in something except a immediately line - something that is curving, in different words - is experiencing a centripetal rigidity. in case you positioned a stone on a string and swing it round your head, the string is exerting a centripetal rigidity on the stone. The Moon is orbiting the Earth - the Earth's gravitational field is exerting a centripetal rigidity on the Moon. A motor vehicle is going round a turn - friction from the line is exerting a centripetal rigidity on the tires of the motorized vehicle.
2016-11-23 15:46:10 · answer #4 · answered by mink 4 · 0⤊ 0⤋
Fh = F cos 60
= 2x9.8 x0.5
= 9.8 N
Normal Force acting will be Fhcos30 (looking from elevation view) = 9.8 x 0.866 = 8.487 N
2006-12-03 21:01:23 · answer #5 · answered by Sid Has 3 · 0⤊ 0⤋