The figure below shows a box of dirty money (mass m1 = 4.0 kg) on a frictionless plane inclined at angle θ1 = 35°. The box is connected via a cord of negligible mass to a box of
laundered money (mass m2 = 2.0 kg) on a frictionless plane inclined at angle θ2 = 55°. The pulley is frictionless and has negligible mass. What is the tension in the cord?
The figure looks like a triangle hill, in which box m1 is on the left side and box m2 is on the right side of the hill.
First: m1=4kg, m2=2kg, T?
Second:
m2gsin(55)-T= m2a
+
m1gsin(35)+T=m1a
----------------
=(m2gsin(55)-m1gsin(35)/(m1+m2)
Third: a=(2kg*9.81m/s^2sin(55)-4kg*9.81m/s^2sin(35))/(2kg+4kg
a=1.109m/s^2
Fourth: plug into equation m1--> T=4kg*9.81m/s^2sin(35)+4kg*(1.109m/s^2)= answer
Answer: 18.2N
love you dr. neutron. your step are so easy to learn physic.
To find the tension in the cord, we can use the principle of conservation of energy. When the boxes move, the potential energy is converted into kinetic energy.
Step 1: Find the gravitational force acting on both boxes.
The gravitational force acting on a box can be calculated using the formula: F = m * g, where m is the mass of the box and g is the acceleration due to gravity (9.8 m/s²).
For m1 (box of dirty money):
Force1 = m1 * g
Force1 = 4.0 kg * 9.8 m/s²
Force1 = 39.2 N
For m2 (box of laundered money):
Force2 = m2 * g
Force2 = 2.0 kg * 9.8 m/s²
Force2 = 19.6 N
Step 2: Find the gravitational components parallel to the inclined planes.
To do this, we need to find the parallel components of the gravitational force acting on each box.
For m1 (box of dirty money):
Force1_parallel = Force1 * sin(θ1)
Force1_parallel = 39.2 N * sin(35°)
Force1_parallel ≈ 22.57 N
For m2 (box of laundered money):
Force2_parallel = Force2 * sin(θ2)
Force2_parallel = 19.6 N * sin(55°)
Force2_parallel ≈ 16.46 N
Step 3: Apply Newton's second law in the direction of motion for each box.
For m1 (box of dirty money):
Sum of forces in the direction of motion = m1 * acceleration1
Force1_parallel - T = m1 * a1
where T is the tension in the cord and a1 is the acceleration of box m1.
For m2 (box of laundered money):
Sum of forces in the direction of motion = m2 * acceleration2
T - Force2_parallel = m2 * a2
where a2 is the acceleration of box m2.
Step 4: Relate the accelerations of the two boxes using the relationship of the inclined planes.
Since the inclined planes are connected by a cord passing over a frictionless pulley, the accelerations of the boxes must be equal in magnitude but opposite in direction.
a1 = -a2 (opposite in direction)
Step 5: Solve the system of equations for T (the tension in the cord).
From Step 3, we have:
-T + 22.57 N = 4.0 kg * (-a2)
From Step 4, we have:
T - 16.46 N = 2.0 kg * a2
Solving these equations simultaneously will give us the value of T (the tension in the cord).
I'll continue with the calculations in the next response.
To find the tension in the cord, we can use the principle of conservation of energy.
First, let's find the gravitational force acting on each box. The force of gravity can be calculated using the equation:
F = m * g
where m is the mass of the box and g is the acceleration due to gravity (approximated as 9.8 m/s²).
For the first box, m1, the gravitational force acting downwards along the incline is:
F1 = m1 * g * sin(θ1)
For the second box, m2, the gravitational force acting downwards along the incline is:
F2 = m2 * g * sin(θ2)
Next, let's consider the motion of the boxes. Since the pulley is frictionless, the tension in the cord will be the same for both boxes.
Let T be the tension in the cord.
For box m1, the net force acting along the incline is the component of the weight in that direction minus the tension in the cord:
m1 * g * sin(θ1) - T = m1 * a
where a is the acceleration of box m1.
For box m2, the net force acting along the incline is the component of the weight in that direction plus the tension in the cord:
m2 * g * sin(θ2) + T = m2 * a
Solving these equations simultaneously will give us the tension in the cord, T.
However, we also need to use the fact that the acceleration of both boxes are equal in magnitude but opposite in direction:
a = -a
This is because, as one box accelerates down the incline, the other box accelerates up the incline at the same rate.
Substituting this expression for a into the equations above, we have:
m1 * g * sin(θ1) - T = -m1 * a
m2 * g * sin(θ2) + T = -m2 * a
We can now solve these equations to find the value of T, the tension in the cord.