Two 10 kilogram boxes are connected by a massless frictionless pulley. The boxes remain st rest with the one on the right hanging vertically and the one on the left 2.0 meters from the bottom of an inclined plane that makes an angle of 60 degrees with the horizontal. The coefficients of kinetic friction and static friction between the left hand box and the plane are 0.15 and 0.30 respectively. You may use g= 10 m/s(2nd power), sin 60 degrees = .087, and cos 60 degrees = 0.50.
What is the tension in the string
What is the force/ forces on the box that is on the plane
You need to break the forces into components along the direction of travel.
First, the hanging box, its weight is mg, downward, in the direction of travel.
Second, the box on the plane. The weigh mg has a component along the plane, mgsinTheta, and a component normal to the plane, mgCosTheta. Friction, then, will be mgCosTheta*mu
Next, the force equation.
Netforce= ma
assume the movement is to the right (clockwise)
mg-mgSinTheta-mgCosTheta*mu= ma
so solve for a. If it is positive, your assumptions about direction (and the direction of friction) was correct.
Next tension. The force on the rope just above the hanging box is
mg-ma. The force on the box on the plane is mgSinTheta+mgCosTheta*mu + ma
check my thinking.
To find the tension in the string, we need to analyze the forces acting on the system. Considering the box on the inclined plane, the normal force (N) will be perpendicular to the inclined surface. The weight of the box (mg) will act straight downward, parallel to the direction of gravity.
Now, let's break down the weight (mg) into its components. The component parallel to the inclined plane is mg*sin(60°), and the component perpendicular to the inclined plane is mg*cos(60°).
The forces acting on the box on the plane are:
1. Weight (mg): The weight acts downward and can be broken down into two components, mg*sin(60°) parallel to the incline, and mg*cos(60°) perpendicular to the incline.
2. Normal force (N): The normal force acts perpendicular to the surface of the incline.
3. Frictional force (Fk): The kinetic friction acts parallel to the incline and opposes the motion. The formula for the kinetic friction is Fk = μk*N, where μk is the coefficient of kinetic friction.
So, the equation for the forces on the box on the plane is:
mg*sin(60°) - Fk - T = 0, where T is the tension in the string.
Next, we can consider the box hanging vertically. The forces acting on the hanging box are:
1. Weight (mg): The weight acts straight downward.
2. Tension in the string (T): The tension in the string acts upward.
Since there is no motion in the vertical direction, the forces balance each other.
The equation for the forces on the hanging box is:
T - mg = 0.
Now, we can solve this system of equations to find the tension in the string:
We know that the weight (mg) is equal to the mass (m) multiplied by the acceleration due to gravity (g). Therefore, mg = 10kg * (10m/s^2) = 100 N.
Substituting this value into the second equation above, we get:
T - 100 N = 0.
So, T = 100 N.
This is the tension in the string.
To find the force/forces on the box that is on the plane, we can substitute the value of T into the first equation:
10kg * (10m/s^2) * sin(60°) - Fk - 100 N = 0.
We know that sin(60°) = 0.87. Also, the coefficient of kinetic friction (μk) is given as 0.15.
Substituting these values, we get:
100 N * 0.87 - (0.15 * N) - 100 N = 0.
Simplifying the equation, we get:
87 N - 0.15N - 100 N = 0.
Combining like terms, we have:
-0.15N - 100 N = -87 N.
Simplifying further, we find:
-0.15N - 100 N = -87 N.
Rearranging the equation, we get:
-0.15N = -13 N.
Dividing both sides by -0.15, we find:
N = 13 N.
So, the force on the box that is on the plane is 13 N.
Therefore, the tension in the string is 100 N, and the force on the box on the plane is 13 N.
To find the tension in the string, we need to calculate the net force acting on the system of the two connected boxes.
Let's start by calculating the weight of each box:
The weight of the box hanging vertically on the right side is given by:
W1 = m1 * g = 10 kg * 10 m/s^2 = 100 N
The weight of the box on the left side can be calculated using the angle of the inclined plane. The component of the weight parallel to the inclined plane is given by:
W2_parallel = m2 * g * sin(60 degrees) = 10 kg * 10 m/s^2 * 0.087 = 8.7 N
The normal force on the box on the left side can be calculated using the angle of the inclined plane. The component of the weight perpendicular to the inclined plane is equal to the normal force:
N = m2 * g * cos(60 degrees) = 10 kg * 10 m/s^2 * 0.5 = 50 N
Now let's calculate the force of friction acting on the box on the left side, which is given by the product of the coefficient of static friction and the normal force:
F_friction_static = μs * N = 0.30 * 50 N = 15 N
The maximum force of static friction possible is equal to the force of friction needed to prevent the box from sliding down the inclined plane. Therefore, the maximum force of static friction is equal to the component of the weight parallel to the plane:
F_friction_static_max = W2_parallel = 8.7 N
Since the force of static friction (15 N) is greater than the maximum force of static friction (8.7 N), the box on the left side will start moving down the inclined plane. This means that the force of friction changes from static friction to kinetic friction.
The force of kinetic friction is given by the product of the coefficient of kinetic friction and the normal force:
F_friction_kinetic = μk * N = 0.15 * 50 N = 7.5 N
Now let's calculate the net force acting on the system:
Net force = W1 - F_friction_kinetic
= 100 N - 7.5 N
= 92.5 N
Since the boxes are connected by a massless and frictionless pulley, the tension in the string will be the same for both boxes. Therefore, the tension in the string is 92.5 N.
Now, let's determine the force(s) acting on the box that is on the inclined plane:
The force of kinetic friction (7.5 N) is acting in the opposite direction to the box's motion down the plane. Therefore, we have:
Force on the box = Net force - F_friction_kinetic
= 92.5 N - 7.5 N
= 85 N
So, the force acting on the box that is on the inclined plane is 85 N.
Wrong...
Your logic is terrible. Fix it.