a box of mass 2.00kg is on a smooth frictionless ramp which makes an angle of 30.0 degree with the horizontal. A force P = 15.0N which makes an angle of 25.0 degree with the vertical is applied to the box. calculate the normal force component that the ramp exerts on the box.
2006-10-24 16:57:09 · 2 answers · asked by Galaxy D 2 in Science & Mathematics ➔ Physics
(2*9.8*cos30 - 15*sin35) N or
(2*9.8*cos30 - 15*sin85) N
You didn't say whether the 15N force was tipped toward the box or away from it
2006-10-24 17:08:24 · answer #1 · answered by Steve 7 · 1⤊ 0⤋
You need to be able to see that the downward force component normal to the ramp exerted by the weight of the box is equal to 19.6cos30. The vertical downward force is 2*9.8=19.6N, 2kg being the mass of the box.
Draw the ramp, the box, and the force P showing the given angles of 30 degrees and 25degrees.
Separately, if your drawing is good, you will see that the force P of 15N exerts and upward force component normal to the ramp. Its direction is (25+30=55) degrees from the line normal to the ramp. This upward component of 15N is equal to 15cos55.
You will note that the force of 19.6cos30 is opposite in direction to 15cos55.
After you have obtained the value of cos55 and cos30 from your table of trigonometric functions, you will then be able to get the values of 19.6cos30 and 15cos55. The normal force component that the ramp exerts on the box will be the difference between 19.6cos30 and 15cos55.
2006-10-25 03:00:55 · answer #2 · answered by tul b 3 · 0⤊ 0⤋