winter 2009, paper 41..
Q4..
(AS level)
Could someone plz include diagrams and explanations in the answer...thanks a lot..
i) First you need to resolve vertically so as to find the reaction.
R = 80 x cos 20 =
75.2 N
Since the block is at rest ----> a = 0
Using Newton's second law of motion : F = ma
T +
Fr - mgsin
x = ma
where
T : tension in the string,
Fr : frictional from between block and the inclined plane
mgsin
x : component of weight down the plane and
x being the angle of inclination of the plane
Given T =13 and a=0 ----> 13 + Fr - 80sin 20 = 0
Hence
Fr is found to be
14.4 N
ii) String is cut off but block is still at rest ----> Fr = mgsin x
Fr = uR where u is the coefficient of friction between block and plane
Hence 80sin 20 = u(75.2) ----->
u = 0.364Sorry I could not add a diagram
Hope it helps