A level Physics Help needed... Watch
This is from OCR Physics A 2011 January Q3 B)ii ,in case you need to check it.
The question is
"The angle between the slope and the horizontal is 7 degrees. The weight of the car is 9.0 x 10^3. The car travels is the slope at a constant speed of 18ms. A resistive force of 300 N acts on the car down the slope.
Calculate the component of the weight of the car acting down the slope"
BTW, the slope is going diagonally upward to the right if that helps.
I drew a vector triangle, so I has 7 degrees as angle bottom left, a 90 degree angle bottom right, and therefore 83 degree angle top right. The vertical component of weight being 9 x 10^3 as opposite side to angle of 7 degrees, and labelled the bit I want (the hypotenuse to the 7 degree angle) as a bold W. Then, basic trig tells me that Sin(7) = opp/hyp = 9x10^3 / W
I rearrange that to W = 9 x 10^3 / sin(7) and get the wrong answer, even after checking ect, please help! Absolutely no idea what I've done wrong, which is why I know I've missed something massive. Thanks in advance!
P.S I keep getting 1.37 x 10^4, the answer is 1.1 x 10^3
Also, if you're able to help, please explain why so I know for future, cheers!
Not sure if this helps you, I am bad at explaining using words.
the top angle is still 7
it's the bottom left is 83
To find the opp, you can use sin, ie sin (opp/hyp)
sin7 x 9x10^3
round it up 1.1x10^3
(ps if im wrong, please correct me)
Sin of the angle = opposite/hypotenuse
The component of weight down the slope is the opposite.
So the answer is mg x sin7