ok. I just did everything again. And I found out that I went wrong with the sins and cos during the calcualtions. My value for R comes out to be 88.3N which is correct.
but the force P (which is unknown at the moment) will also need to be taken into consideration. If I resolved into perpendicular components it becomes Pcos60 +6gcos30 = R?
Another problem I can't solve
A box of mass 6kg lies on a rough plane inclined at an angle of 30 to the horizontal. The box is held in equilibrium by means of a horizontal force of magnitude P Newtons.
The coefficent of friction between the box and the plane is 0.4.
Find the normal...
Hi,
has anyone heard the song FOREVER by Jessy and the Rippers. It is such an amazing song but so difficult to find.
Can anyone help me find it? :rolleyes:
The initial speed of a car is 25ms^-1 and it accelerates for 4 secs to reach a speed of Vms^-1. It then maintains this velocity traveling at constant speed for a further 8 seconds. The total overla distance travlled is 600m.
I am really stuck.
I used v = u+at to get one question.
v = 25 +...
Oh no! Did I just make the mistake of the century? This is soo embarassing. OBVIOuSLY 2T/T is not T!...SHOOT ME...and yes, what you said is right. How could I forget these simple rules? all right..let me try again
how? that doesn't make sense.
Tcos60 = 2Tcos (90-θ)
since cos 90 = 0, the equation simplifies to: Tcos60 = -2Tcos θ
cancel one T on each side>>>> cos60 = -tcosθ Now??