Mechanics question on forces

[weak locust]

tan^-1(3/4) gives angle alpha which is 36.87 degrees

if you let alpha equal theta, and mass 0.5 = m, then you can apply this diagram

since the particle is held at rest, the forces must be balanced

so normal reaction force = mgcos(theta) = force the weight exerts on the slope

when you plug in all the numbers ||((9.81)(0.5)cos(36.87))|| you get 3.92

out of interest, was this for physics or maths?

[weak locust]

ah right, in that case i think you're supposed to use 9.8 instead of 9.81 for g

but the answer should still be similar

no problems

[sand hatch]

it's year 1 mechanics

year 2 is moments, projectiles, coefficient of restitution

[weak locust]

it's definitely year 2

ive finished all of year 1 and inclined planes havent come up

[sand hatch]

i did it year 1

it's in M1

on the old spec

which is year 1

[weak locust]

tbf content order may have changed from old spec to new

cuz in the new spec it's year 2

[sand hatch]

it's really easy

it's just some resolving

[weak locust]

yeah, it gets taught in year 1 physics anyway

[sand hatch]

also don't do arctan

make a triangle

to get exact sin and cos values

not approximations

or use trig identities

never arctan tho

[weak locust]

as in this?

im not sure how you would do it without arctan tbh

never done it a different way

[sand hatch]

tan is opposite over adjacent yeh

so draw a right angled triangle

if tan alpha is 3/4, then proportionally opposite = 3, adjacent = 4 and by using pythagoras obviously hypotenuse is 5

[weak locust]

ah i see

[sand hatch]

and then cos is adjacent over hypotenuse (4/5), and sin is opposite over hypotenuse (3/5)

easy

[weak locust]

cool

[sand hatch]

or use trig identities

sec^2 alpha is tan^2 alpha + 1

so sec^2 alpha is 25/16

so cos^2 alpha is 16/25

so cos alpha is +_ 4/5

but as cos is in this case positive, cos alpha must be 4/5

[weak locust]

i havent started year 13 maths yet so i havent learnt that

but i think i get it

sec is 1/cos right