A particle P of mass 2 kg is held at rest in equilibrium on a rough plan. The plane is inclined to the horizontal at an angle of 20°. Find the coefficient of friction between P and the plane.
For any mechanics question such as this one the best way to start off is by drawing a diagram of the system and including all the forces acting on the particle P. In order to do so, we have to account for three forces: the gravitational force which always acts directly downwards (i.e. towards the centre of the Earth). Its given by mg where m is the particle's mass and g is the gravitational acceleration (9.81 m/s2). the normal force N which always acts in order to keep the particle in the plane by opposing gravity's vertical component (an object standing on a flat table will not plunge into the floor due to gravity as the normal force keeps it in the table).the frictional force FRwhich always acts parallel to the plane and opposing the particle's direction of motion (which in this case is down the slope due to gravity). Its related to N by the equation: FR = 'mu '. N , where 'mu' is known as the coefficient of friction which indicates how rough the plane's surface is. Then one must use the given state of the particle, in this case as its in equilibrium (no acceleration) the forces acting on P must be balanced. For convenience we have set the x and y axes parallel/perpendicular to the plane which allows us to decompose the gravitational force into an horizontal and a vertical component which directly oppose the frictional and normal forces, respectively. Using the plane's angle with the horizontal ( 'theta' ) and some simple trigonometry, it can be shown that the vertical component is given by mgcos('theta') (the closest component to the total force is always the one with a cosine) and similarly for the horizontal component.Now its just a matter of balancing forces by setting the total force of each axis to zero: Fy= N - mgcos(theta) = 0 ; and Fx = mgsin(theta) - FR = 0Putting in the values given in the question and the known expression for friction, we get:N = 2gcos(20º) ; and 2gsin(20º) = (mu).N >>>> 2gsin(20º) = mu . 2gcos(20º)Solving for the unknown (which is mu), we get the result:mu = sin(20º) / cos(20º) = 0.364 (to 3 significant figures)* A good way to know if your value for mu is correct is that 'mu' must always be a value between 0 and 1.
