Two wooden crates with masses are as shown are tied together by a horizontal cord. Another cord is tied to the first crate and it is pulled with a force of 199N at a angle of 20.0 degrees. Each crate has a coefficient of kinetic friction of 0.55 . Find the tension in the ropes between the crates and the magnitude of the acceleration of the system.
The smallest mass is 9.91kg which is attached to the biggest mass of 11.2kg. and the rope is pulled at 20.0 degrees with a force of 199N...
To find the tension in the ropes between the crates and the magnitude of the acceleration of the system, you can use Newton's second law of motion.
Step 1: Resolve the force into its horizontal and vertical components.
The force applied to the first crate can be resolved into its horizontal and vertical components by using trigonometry. The horizontal component can be found by multiplying the force by the cosine of the angle, and the vertical component can be found by multiplying the force by the sine of the angle.
Horizontal component: F_horizontal = 199N * cos(20°)
Vertical component: F_vertical = 199N * sin(20°)
Step 2: Calculate the force of friction on each crate.
The force of friction can be calculated by multiplying the coefficient of kinetic friction by the normal force acting on each crate. In this case, the normal force is equal to the weight (mass * gravity).
Force of friction on the first crate: F_friction1 = coefficient_of_friction * mass1 * gravity
Force of friction on the second crate: F_friction2 = coefficient_of_friction * mass2 * gravity
Step 3: Apply Newton's second law of motion.
The net force acting on each crate is equal to the force applied minus the force of friction.
Net force on the first crate: F_net1 = F_horizontal - F_friction1
Net force on the second crate: F_net2 = F_friction1 - F_friction2
Since the two crates are tied together by a horizontal cord and experience the same acceleration, the net forces on the crates will be equal.
F_net1 = F_net2
Step 4: Determine the tension in the ropes.
The tension in the rope between the crates is equal to the net force acting on each crate.
Tension in the ropes: T = F_net1
Step 5: Calculate the acceleration of the system.
The acceleration can be found by dividing the net force by the total mass of the system.
Acceleration of the system: a = F_net1 / (mass1 + mass2)
Now you can substitute the known values into the equations to find the tension in the ropes and the magnitude of the acceleration.
To solve this problem, we can break it down into steps:
Step 1: Find the force due to friction on each crate.
The force of kinetic friction on each crate can be calculated using the formula:
Frictional force = coefficient of kinetic friction * normal force
The normal force on each crate is equal to the weight of the crate, which can be calculated as:
Normal force = mass of crate * acceleration due to gravity
Mass of larger crate = 11.2 kg
Mass of smaller crate = 9.91 kg
Acceleration due to gravity = 9.8 m/s^2
For the larger crate:
Normal force (N1) = Mass of larger crate * Acceleration due to gravity
= 11.2 kg * 9.8 m/s^2
For the smaller crate:
Normal force (N2) = Mass of smaller crate * Acceleration due to gravity
= 9.91 kg * 9.8 m/s^2
Step 2: Find the tension in the ropes between the crates.
Tension in the rope connecting the crates (T1) is equal to the force exerted on the larger crate minus the frictional force on the larger crate.
T1 = Force on larger crate - Frictional force on larger crate
= Force on larger crate - coefficient of kinetic friction * normal force on larger crate
Similarly, tension in the other rope (T2) is equal to the force exerted on the smaller crate minus the frictional force on the smaller crate.
T2 = Force on smaller crate - Frictional force on smaller crate
= Force on smaller crate - coefficient of kinetic friction * normal force on smaller crate
Step 3: Calculate the force exerted on each crate.
The force exerted on each crate can be determined by resolving the applied force into horizontal and vertical components.
Force applied on the system (Fapplied) = 199 N
Angle of applied force (θ) = 20.0 degrees
Force exerted on the larger crate (F1) = Fapplied * cos(θ)
Force exerted on the smaller crate (F2) = Fapplied * sin(θ)
Step 4: Calculate the acceleration of the system.
Since the two crates are tied together, they have the same acceleration. We can calculate the acceleration using Newton's second law:
Net force on the system = (Mass of larger crate + Mass of smaller crate) * Acceleration
Net force on the system = (Force exerted on larger crate - Frictional force on larger crate) + (Force exerted on smaller crate - Frictional force on smaller crate)
Solve this equation for acceleration:
(Mass of larger crate + Mass of smaller crate) * Acceleration = (Force exerted on larger crate - Frictional force on larger crate) + (Force exerted on smaller crate - Frictional force on smaller crate)
Step 5: Substitute the known values to find the answers.
Using the given values, substitute them into the appropriate equations to find the values of T1 (tension in the rope between the crates), T2 (tension in the other rope), and the acceleration of the system.