A rubber chicken of mass 45.0 kg is pulled along a horizontal surface by a rope that makes an angle of 32.0 with the horizontal. If the coefficient of friction is 0.113 and the tension in the rope is 95.0 N, what is the acceleration of the object?
X: ma=Fcosα-F(fr)
Y: 0=-mg+N+Fsinα =>N=mg-Fsinα
F(fr) = μN=μ(mg-Fsinα)
ma= Fcosα- μ(mg-Fsinα)
a = { Fcosα- μ(mg-Fsinα)}/m
To find the acceleration of the object, we need to consider the forces acting on it.
First, let's analyze the forces in the horizontal direction.
The tension in the rope has a horizontal component given by T * cos(θ), where T is the tension (95.0 N) and θ is the angle of the rope (32.0 degrees).
The friction force opposing the motion is given by the coefficient of friction (μ) multiplied by the normal force (N). The normal force is equal to the weight of the rubber chicken, which is given by mass (m) multiplied by the acceleration due to gravity (g).
In this case, g is approximately 9.8 m/s^2. So, the normal force is N = m * g = 45.0 kg * 9.8 m/s^2.
The friction force is then Friction = μ * N = 0.113 * (45.0 kg * 9.8 m/s^2).
Since the object is being pulled, the net force acting in the horizontal direction is F_net = T * cos(θ) - Friction.
Using Newton's second law, F_net = m * a, where m is the mass (45.0 kg) and a is the acceleration we are trying to find.
Combining the equations, we have m * a = T * cos(θ) - Friction.
Now we can calculate the acceleration of the object.
To find the acceleration of the object, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration.
First, we need to find the net force acting on the object. There are two forces acting on the object: the tension force in the rope and the force of friction.
The tension force can be calculated using trigonometry. The vertical component of the tension force is T * sin(θ), and the horizontal component is T * cos(θ), where T is the tension in the rope and θ is the angle it makes with the horizontal. In this case, T = 95.0 N and θ = 32.0°.
Vertical component of tension force:
T * sin(θ) = 95.0 N * sin(32.0°) = 49.5 N
Horizontal component of tension force:
T * cos(θ) = 95.0 N * cos(32.0°) = 80.6 N
The force of friction can be calculated using the equation F_friction = μ * N, where μ is the coefficient of friction and N is the normal force. The normal force is equal to the gravitational force acting on the object, which is given by N = m * g, where m is the mass of the object and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Mass of the object: 45.0 kg
Normal force: m * g = 45.0 kg * 9.8 m/s^2 = 441 N
Force of friction:
F_friction = μ * N = 0.113 * 441 N = 49.833 N
Now we can calculate the net force by subtracting the force of friction from the horizontal component of tension force:
Net force = (T * cos(θ)) - F_friction = 80.6 N - 49.833 N = 30.767 N
Finally, we can use Newton's second law to find the acceleration:
Net force = mass * acceleration
30.767 N = 45.0 kg * acceleration
Solving for acceleration:
acceleration = 30.767 N / 45.0 kg = 0.683 m/s^2
Therefore, the acceleration of the object is 0.683 m/s^2.