A particle of mass 50kg is placed on a ramp at 30 degrees, forces acting on it are Weight, Reaction and Friction, the weight (previously figured out is 490N, how can I calculate the component of W which acts down the slope?
2007-02-16 00:35:53 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Physics
reaction force or N is :
mgcos30
and component of weight which acts down the slope is :
mgSin30=50x0.5=25
2007-02-16 01:05:22 · answer #1 · answered by saleh_sepehri 2 · 0⤊ 0⤋
Newton's do-not come in to the answer
It is triangle of forces. The included angle is 60 not 30 degrees.
i.e. the force is split into 2 directions the hypotenuse being the longest must be the greater force
Therefore friction =50 X 1.732 /2 = 43.3Kg or
friction = 50 X Sine60
=50 X 0.866 =43.3Kg
=force acting down the slope
The reactive force acting upwards
=50 X1/2 =25Kg or
=50 X cos 60
=50 X 0.5 =25Kg
2007-02-20 00:00:19 · answer #2 · answered by mad_jim 3 · 0⤊ 0⤋
490 * cos 30
2007-02-16 01:06:08 · answer #3 · answered by koki83 4 · 0⤊ 0⤋
just multyply 490N with sin30.u can get the answer
2007-02-16 01:06:14 · answer #4 · answered by Anonymous · 0⤊ 0⤋
Are you askimg me to do your homework?
2007-02-16 10:45:18 · answer #5 · answered by Pete WG 4 · 0⤊ 0⤋