The frictional force experienced by a body placed on an inclined plane as shown in the diagram below is:
$$\text{frictional force} = μR$$ $$\text{Normal reaction} = mgcosθ $$ $$\text{frictional force} = μMgcosθ $$ $$\text{the force against against friction} $$ $$ f = mgsinθ $$ $$\text {Net force, f} = mgsinθ - μmgcosθ $$ $$\text{Net force, f} = mgsinθ - F $$ $$ where,\text{ F = Frictional force }$$ $$\text{To get the Value for acceleration} $$ $$ F = ma $$ $$ ma =mg(sinθ - μcosθ) $$ $$\text{acceleration, a} = g(sinθ - μcosθ) $$ Example 1 : A concrete block of mass 25kg is placed on a wooden plank inclined at an angle of 60° to the horizontal. Calculate the force parallel to the inclined plane that will keep the block at rest if the coefficient of friction between the block and the plank is 0.45
Solution
Example 2: The diagram above shows a body resting on an inclined plane. If the body slides down the plane, what will be its acceleration? (g = 10m/s²) (WAEC)
Solution
Example 3 : Two masses M1 = 5kg and M2 = 10kg are connected at the ends of an inextensible string passing over a frictionless pulley as shown. When the masses are released, then the acceleration of the masses will be:
Solution
$$\text{For }{M_2} $$ $$\text{Tension in the string} = mg - ma $$ $$ T = {10}g - {10}a $$ $$ \text{equating both equations} $$ $$ 5a + 5g = 10g - 10a $$ $$ 5a + 10a = 10g - 5g $$ $$ 15a = 5g $$ $$ a = \frac{5g}{15} = \frac{g}{3} $$
Example 4 : The acceleration of the system shown below is? (JAMB)
Solution
$$\text{for the 30kg mass} $$ $$ T = mg - ma $$ $$ T = 30 × 10 - 30a $$ $$ T = 300 - 30a $$ $$\text{equating both} $$ $$20a + 100 = 300 - 30a $$ $$ 20a + 30a = 300 - 100 $$ $$ 50a = 200 $$ $$ a = \frac{200}{50} $$ $$ a = 4m/s² $$
Example 5 : A motorcycle of mass 100kg moves round in a circle of radius 10m with a velocity of 5m/s. Find the coefficient of friction between the road and tyres.