Two blocks with equal masses m1 = m2
3.2 kg are connected via a pulley as shown in the above figure. Take the coefficent of kinetic friction to be
0.3 and the angle of the incline to be 40 degrees.
Find the acceleration given that
(Note: enter a negative value if the acceleration on block 1 is directed downwards.)
a) m1 is moving down
b) m1 is moving up
c) If m2 = 6.2 kg, for what values of m1 will the pair move at constant speed?
To find the acceleration, we need to consider the forces acting on each block and apply Newton's laws of motion.
Let's start by analyzing the forces acting on each block individually:
For Block m1:
1. Weight (mg1): The weight of m1 is given by mg1, where g is the acceleration due to gravity.
2. Normal force (N1): The normal force is the force exerted by the incline on m1 and is perpendicular to the surface of the incline.
3. Friction force (f1): The friction force opposes the motion of m1 and can be calculated using the coefficient of kinetic friction and the normal force.
For Block m2:
1. Weight (mg2): The weight of m2 is given by mg2.
2. Normal force (N2): The normal force is the force exerted by the pulley on m2.
Now, let's consider the tensions in the string:
1. Tension T1: The tension in the string acting on m1 and directed upwards.
2. Tension T2: The tension in the string acting on m2 and directed downwards.
Forces on m1 along the incline:
1. The weight component (mg1*sinθ) parallel to the incline, acting downwards.
2. The tension T1 acting upwards.
3. The friction force f1 acting downwards.
The net force on m1 along the incline is given by:
Net force = (mg1*sinθ) - T1 - f1
Forces on m2 vertically:
1. The weight component (mg2*cosθ) acting downwards.
2. The tension T2 acting upwards.
The net force on m2 vertically is given by:
Net force = T2 - (mg2*cosθ)
Taking the positive direction as up the incline and up vertically, we can write the equations of motion for m1 and m2:
For m1:
(mg1*sinθ) - T1 - f1 = m1*a (equation 1)
For m2:
T2 - (mg2*cosθ) = m2*a (equation 2)
Now, let's solve the equations to find the acceleration:
Substituting f1 = μ*N1 into equation 1, where μ is the coefficient of kinetic friction and N1 is the normal force on m1:
(mg1*sinθ) - T1 - μ*N1 = m1*a
Substituting N1 = m1*g*cosθ into the above equation:
(mg1*sinθ) - T1 - μ*m1*g*cosθ = m1*a
Rearranging the equation and substituting T2 to eliminate T1:
(mg1*sinθ) - μ*m1*g*cosθ - T2 = m1*a (equation 3)
Rearranging equation 2 to express T2 in terms of T1:
T2 = (mg2*cosθ) + m2*a
Substituting T2 into equation 3 and rearranging:
(mg1*sinθ) - μ*m1*g*cosθ - (mg2*cosθ) - m2*a = m1*a
Simplifying the equation:
(mg1*sinθ) - (mg2*cosθ) = (m1 + m2 + μ*m1)*a
Finally, we can solve for the acceleration (a) using the given values of m1, m2, θ, and μ.
For part a) when m1 is moving down, the acceleration will be negative.
For part b) when m1 is moving up, the acceleration will be positive.
For part c) when the pair moves at constant speed, the net force on both blocks will be zero. So, the equation can be written as:
(mg1*sinθ) - (mg2*cosθ) = 0
Solving for m1 using the given value of m2 and the other known quantities, we can find the range of values for m1 where the pair moves at constant speed.
To find the acceleration in each scenario, we can use Newton's laws of motion and consider the forces acting on each block.
a) When m1 is moving down:
In this case, the net force on m1 will be downwards. To find the acceleration, we first need to calculate the net force on m1.
For m1:
The gravitational force acting on m1 is given by Fg1 = m1 * g, where g is the acceleration due to gravity.
The frictional force acting on m1 is given by Ff = μ * m1 * g * cos(θ), where μ is the coefficient of kinetic friction and θ is the angle of the incline.
The net force on m1 is given by Fnet1 = Fg1 - Ff.
Since m1 is moving downwards, the direction of the net force will also be downwards, so we can write Fnet1 = m1 * a, where a is the acceleration of m1.
Combining the above equations, we can write:
m1 * a = m1 * g - μ * m1 * g * cos(θ)
Simplifying the equation, we get:
a = g - μ * g * cos(θ)
Substituting the given values:
m1 = m2 = 3.2 kg
μ = 0.3
θ = 40 degrees
and using the standard value for the acceleration due to gravity, g = 9.8 m/s^2, we can calculate the acceleration.
a = 9.8 - 0.3 * 9.8 * cos(40)
= 9.8 - 0.3 * 9.8 * 0.766
= 9.8 - 2.261
≈ 7.54 m/s^2 (downwards)
Therefore, the acceleration of m1 when it is moving down is approximately 7.54 m/s^2 downwards.
b) When m1 is moving up:
Similar to the previous case, the net force on m1 will be upwards.
Using the same equations, the acceleration when m1 is moving up is given by:
a = g + μ * g * cos(θ)
Substituting the given values, we get:
a = 9.8 + 0.3 * 9.8 * 0.766
≈ 9.8 + 2.261
≈ 12.06 m/s^2 (upwards)
Therefore, the acceleration of m1 when it is moving up is approximately 12.06 m/s^2 upwards.
c) For what values of m1 will the pair move at constant speed?
When the pair moves at constant speed, the net force on the system is zero.
For m1:
The gravitational force acting on m1 is given by Fg1 = m1 * g.
The frictional force acting on m1 is given by Ff = μ * m1 * g * cos(θ).
The tension in the string is the same for both blocks and given by Ft = m2 * g.
Since the pair is moving at constant speed, the net force on m1 is zero, so we can write:
Fnet1 = Fg1 - Ff - Ft = 0
m1 * g - μ * m1 * g * cos(θ) - m2 * g = 0
Solving for m1, we get:
m1 = m2 / (1 - μ * cos(θ))
Substituting the given values:
m2 = 6.2 kg
μ = 0.3
θ = 40 degrees
m1 = 6.2 / (1 - 0.3 * cos(40))
≈ 6.2 / (1 - 0.3 * 0.766)
≈ 6.2 / (1 - 0.2298)
≈ 6.2 / 0.7702
≈ 8.04 kg
Therefore, for m1 to move at a constant speed with m2 = 6.2 kg, the value of m1 should be approximately 8.04 kg.