A 27.9 kg block (m1) is on a horizontal surface, connected to a 6.50 kg block (m2) by a massless string as shown in the figure below. The frictionless pulley has a radius R = 0.090 m and a moment of inertia I=0.060 kgm2. A force F = 192.1 N acts on m1 at an angle theta = 32.1°. There is no friction between m1 and the surface. What is the upward acceleration of m2?
To find the upward acceleration of m2, we can use Newton's laws of motion and consider the forces acting on the system.
Let's break down the forces acting on each block:
For m1:
1. The force F acting at an angle θ = 32.1°, which can be resolved into its horizontal and vertical components.
F_h = F * cos(θ)
F_v = F * sin(θ)
2. The weight of m1, which acts vertically downward.
W1 = m1 * g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
For m2:
1. The weight of m2, which acts vertically downward.
W2 = m2 * g
Since the pulley is frictionless, we don't have to consider any rotational forces or torque acting on it.
Now, let's consider the tension in the string. The tension will be the same throughout the string.
Using Newton's second law (F = ma), we can write the equations of motion for the system:
For m1:
1. In the horizontal direction:
F_h - T = m1 * a, where a is the common acceleration of both blocks.
2. In the vertical direction:
T - W1 = 0 (since there is no vertical acceleration)
For m2:
1. In the vertical direction:
W2 - T = m2 * a
We can now solve these equations simultaneously to find the value of the common acceleration (a).
1. Start with the equation for m2 in the vertical direction:
W2 - T = m2 * a
2. From the equation for m1 in the horizontal direction:
F_h - T = m1 * a
3. Substitute T from equation 2 into equation 1:
W2 - (F_h - m1 * a) = m2 * a
4. Rearrange the equation:
W2 - F_h = (m1 + m2) * a
5. Substitute the values:
W2 = m2 * g
F_h = F * cos(θ)
(m2 * g) - (F * cos(θ)) = (m1 + m2) * a
6. Solve for a:
a = [(m2 * g) - (F * cos(θ))] / (m1 + m2)
Now we can substitute the given values to find the upward acceleration of m2.