A 3.3 kg wooden block is placed on 22 degree inclined plane and is connected by a string that passes over a light friciton less pulley at the top of the incline and then attached to a hanging 3.0kg block.If the coefficient of static friction bet the 1st block and the surface of the plane is 0.25 and coefficient of kinetic friction is 0.15
a)which direction will the first block move, up the plane or down the plane if it does.
b)Calculate the acceleration of the system and the tension force in the string.
Draw the diagram. Next resolve the weight of the block into normal, and down the plane components.
Normal: mgCosTheta
downtheplane: mgSinTheta
Up the plane force is 3g
Friciton is mgCosTheta*mu
Now sum forces: Assume friction is down the plane.
Net force= totalmass*a
3g-mgSinTheta-mgCosTheta*mu= (3+3.3)a
solve for a. If a is zero, then you chose the wrong direction for friction, and you know answer a).
A doubly tethered model airplane of mass 1.0 kg is attached to a vertical rod by two very light strings that are 2.0 each in length. the airplane and the the strings rotate in a horizontal circleabout the rod with both the strings remaining taut.The attachment points of the strings are also separated by 2.0 m If the tension in the upper string is 39.2 N. Calculate
a)the tension in the lower string.Is it likely to be less or more than the tension in the upper string?
b)the steady speed of the ball as it undergoes circular motion.
To determine the direction of the first block's movement, we need to compare the forces acting on it.
First, let's calculate the normal force (N) acting on the first block. The normal force is equal to the component of the weight perpendicular to the inclined plane.
N = m1 * g * cos(θ)
where m1 is the mass of the first block, g is the acceleration due to gravity, and θ is the angle of the inclined plane (22 degrees).
N = 3.3 kg * 9.8 m/s^2 * cos(22°)
N ≈ 28.07 N
The force of gravity acting parallel to the inclined plane (F_parallel) is given by:
F_parallel = m1 * g * sin(θ)
F_parallel = 3.3 kg * 9.8 m/s^2 * sin(22°)
F_parallel ≈ 13.64 N
Next, let's determine the maximum static friction force (F_static_max) between the first block and the inclined plane. The maximum static friction force is given by:
F_static_max = μ_s * N
where μ_s is the coefficient of static friction. In this case, μ_s = 0.25.
F_static_max = 0.25 * 28.07 N
F_static_max ≈ 7.02 N
Since the force of gravity parallel to the incline (F_parallel = 13.64 N) is greater than the maximum static friction force (F_static_max = 7.02 N), the block will overcome static friction and start moving down the plane.
Now, let's calculate the acceleration of the system and the tension force in the string.
The net force (F_net) acting on the system is given by:
F_net = m1 * a
where m1 is the mass of the first block and a is its acceleration.
The force of gravity acting on the hanging block (m2) is given by:
F_gravity = m2 * g
F_gravity = 3.0 kg * 9.8 m/s^2
F_gravity ≈ 29.4 N
The tension force in the string (T) is the force required to accelerate both blocks together. It is also equal to the force of gravity acting on the hanging block.
T = F_gravity = 29.4 N
Since the mass of the two blocks is the same (m1 = 3.3 kg and m2 = 3.0 kg), the net force (F_net) can be written as:
F_net = (m1 + m2) * a
F_net = (3.3 kg + 3.0 kg) * a
F_net ≈ 6.3 kg * a
Since the net force (F_net) is equal to the difference between the force of gravity acting parallel to the inclined plane (F_parallel) and the force of kinetic friction (F_kinetic), we can write:
F_net = F_parallel - F_kinetic
F_net = 13.64 N - F_kinetic
The force of kinetic friction (F_kinetic) is given by:
F_kinetic = μ_k * N
where μ_k is the coefficient of kinetic friction. In this case, μ_k = 0.15.
F_kinetic = 0.15 * 28.07 N
F_kinetic ≈ 4.21 N
Substituting the values, we have:
6.3 kg * a = 13.64 N - 4.21 N
6.3 kg * a ≈ 9.43 N
a ≈ 1.49 m/s^2
The acceleration of the system is approximately 1.49 m/s^2, and the tension force in the string is 29.4 N.