Principles of Physics Week 06 Exercises

other block goes up the ramp ( m1 ), because they are connected by a rope. Assuming no pulley weight or resistance, the tension of the rope is the same on both sides. The total force going down the rope towards m2 is equal to: ( (2M)*g) - (M sin (40)* g ) . These two forces work opposite each other, because the pulley changes the direction of forces that are on the rope. We then need to calculate the force of friction which is F friction = μ ⋅ N. (N=normal force) Since the only source of friction is M, we only need the Normal force of M which is: M*cos(40) . Then we multiply by the coefficient of friction, μ k = 0.40. The final equation we get is : (2M)*g - (M sin (40)* g ) - (M*cos(40)*(0.40)).=F sum By plugging in 1kg as our mass, we find that F sum is Fsum ≈ 12.523 N Given F = m ⋅ a 12.523 N = 3 7) The two blocks shown are released from rest. If the coefficient of kinetic friction between the blocks and the surfaces on which they slide is μ = 0.20, what is the magnitude of the acceleration of either block? Let's start with the horizontal block (Bh) with M (mass). To overcome friction a force of F friction = μ ⋅ N Is needed. The force pushing down on Bh is F =( M ⋅ a) = M ⋅ 9.81 m / s^ 2 And μ is 0.20. So the F friction is M (9.81 m / s^ 2 ) ⋅ ( 0.20 ) = M(1.962N) Now the block on the slope (Bs) with 2M (mass) at θ (theta) = 30 degrees. F horizontal = P ⋅ cos(30.0°) = P (0.866) F vertical = P ⋅ sin(30.0°) = P (0.5) Without friction the acceleration (a) of Bs would be a=m ⋅ g ⋅ sin(30.0°)/m (note the mass cancels )and g=9.81 m / s^ 2 a= (9.81 m / s^ 2 ) ⋅ ( 0.5 ) = 4.905 m / s^ 2 Now add in the friction which is determined by F friction = μ ⋅ N And acceleration is a=g(sin θ - μ cos θ) a= g(.5 ⋅ (.2 ⋅ 0.866)) = g(0.08660) = 0.8495 8) A block of mass 2.50 kg is pushed against a wall by a force P that makes a 50.0° angle with the horizontal. The coefficient of static friction between the block and the wall is 0.300. Find the magnitude of the force P that allows the block to remain stationary and not slide down the wall. Let’s start with the force pulling the block down F =( m ⋅ a) = ( 2.50 kg) ⋅ 9.81 m / s^ 2 = 24.52N