The diagram for this problem is a pulley. On the left end Box 1 is sitting on a table and Box 2 is attached sitting on top of Box 1 and is attached to the string for the pulley. The right side has the string attached to Box 3 which is lifted off the table.
A massless rope passes over a massless, frictionless pulley. One end of the rope is connected to box 2 and the other end is connected to box 3. The masses of the three boxes are
Box 1: 5.4 kg
Box2: 3.7 kg
Box 3: 2.6 kg
Determine the magnitude of the normal force that the table exerts on box 1.
I drew free body diagrams to get me started but can't get much farther than that
To solve this problem, you need to analyze the forces acting on each box and apply Newton's second law of motion.
Let's start by drawing the free body diagrams for each box:
Box 1:
- There are two forces acting on Box 1: the weight (mg) downwards and the normal force (N) upwards. The normal force is the force exerted by the table on Box 1.
- Since Box 1 is on a table, it is not moving vertically, so the net force in the vertical direction is zero. Therefore, the magnitude of the normal force (N) is equal to the weight (mg) acting downwards.
- The mass of Box 1 is given as 5.4 kg, so its weight is (5.4 kg) × (9.8 m/s^2) = 52.92 N.
Box 2:
- There are three forces acting on Box 2: the tension (T) upwards due to the string, the weight (mg) downwards, and the normal force (N) from Box 1.
- Since Box 2 is on top of Box 1, the normal force (N) exerted by Box 1 on Box 2 is equal in magnitude but opposite in direction to the normal force exerted by the table on Box 1.
- The weight of Box 2 is given as 3.7 kg, so its weight is (3.7 kg) × (9.8 m/s^2) = 36.26 N.
- The net force in the vertical direction for Box 2 is the difference between the tension (T) and the weight (mg): T - mg. Since it is not moving vertically, the net force is zero. Therefore, we have T - mg = 0.
Box 3:
- There are two forces acting on Box 3: the tension (T) upwards due to the string and the weight (mg) downwards.
- The weight of Box 3 is given as 2.6 kg, so its weight is (2.6 kg) × (9.8 m/s^2) = 25.48 N.
- The net force in the vertical direction for Box 3 is the difference between the tension (T) and the weight (mg): T - mg. This net force is responsible for lifting Box 3 off the table.
You have a system of equations:
Equation 1: N = 52.92 N (from Box 1)
Equation 2: T - 3.7(9.8) = 0 (from Box 2)
Equation 3: T - 2.6(9.8) = 0 (from Box 3)
Solving this system of equations will give you the tension in the string (T), which is equal to the magnitude of the normal force (N) exerted by the table on Box 1.
Solving the equations yields:
T = 3.7(9.8) = 36.26 N (from Equation 2 and Box 2)
T = 2.6(9.8) = 25.48 N (from Equation 3 and Box 3)
Since the tension in the string is the same at both ends, we can conclude that the magnitude of the normal force (N) exerted by the table on Box 1 is equal to 36.26 N.