Mass 1 (10 kg) rests on a table connected by a string to Mass 2 (5 kg). If s = 0.30 and k = 0.20, what is
a) the acceleration of each block? (Draw the figure and proceed).
b) the tension in the connecting string?
a) To find the acceleration of each block, we can use Newton's second law of motion. We'll consider the forces acting on each block separately.
Block 1 (10 kg):
- There is a tension force pulling it to the right (Tension 1).
- There is a friction force opposing its motion (Friction 1).
- The weight of the block acts downward (Weight 1).
- The normal force from the table acts upward (Normal force 1).
Block 2 (5 kg):
- There is a tension force pulling it to the left (Tension 2).
- There is a friction force opposing its motion (Friction 2).
- The weight of the block acts downward (Weight 2).
- The normal force from the table acts upward (Normal force 2).
Considering friction:
- The maximum static friction force (Fs) is given by Fs = μs * Normal force. So, Friction 1 = μs * Normal force 1 = 0.30 * Normal force 1.
- The kinetic friction force (Fk) is given by Fk = μk * Normal force. So, Friction 2 = μk * Normal force 2 = 0.20 * Normal force 2.
Since both blocks are connected, the tension force (Tension 1) in the connecting string is the same as the tension force (Tension 2) in the other end of the string.
Now, let's proceed with drawing the free-body diagrams for each block and solving for the acceleration:
Block 1 (10 kg):
- The net force in the horizontal direction is Tension 1 - Friction 1 = 10 * acceleration. (Newton's second law)
- The net force in the vertical direction is Normal force 1 - Weight 1 = 0. (Since block 1 is not moving vertically)
- The normal force is equal in magnitude and opposite in direction to the weight, so Normal force 1 = Weight 1 = 10 * 9.8 = 98 N.
Block 2 (5 kg):
- The net force in the horizontal direction is Tension 2 - Friction 2 = 5 * acceleration. (Newton's second law)
- The net force in the vertical direction is Normal force 2 - Weight 2 = 0. (Since block 2 is not moving vertically)
- The normal force is equal in magnitude and opposite in direction to the weight, so Normal force 2 = Weight 2 = 5 * 9.8 = 49 N.
Now, let's solve the equations:
For Block 1: Tension 1 - Friction 1 = 10 * acceleration.
For Block 2: Tension 2 - Friction 2 = 5 * acceleration.
Since Tension 1 = Tension 2, we can simplify the equations:
Tension - 0.30 * Normal force 1 = 10 * acceleration
Tension - 0.20 * Normal force 2 = 5 * acceleration
Substituting the values for the Normal forces, we have:
Tension - 0.30 * 98 = 10 * acceleration
Tension - 0.20 * 49 = 5 * acceleration
Now, we have two equations with two unknowns (Tension and acceleration), so we can solve for them simultaneously.
b) To find the tension in the connecting string, we need to solve one of the above equations for Tension. Let's use the first equation:
Tension = 10 * acceleration + 0.30 * 98
Once we have the value of acceleration from part (a), we can substitute it into this equation to find the tension.
a) To find the acceleration of each block, we need to consider the forces acting on them.
First, let's draw the figure:
_________
| |
M1 | |
________|_________|
| |
| |
| M2 |
|__________________|
The frictional force, denoted by F_friction, opposes the motion of the blocks.
For Mass 1 (M1):
The weight of M1 is given by F_weight_M1 = M1 * g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
The frictional force acting on M1 is given by F_friction_M1 = μs * F_normal_M1, where μs is the coefficient of static friction and F_normal_M1 is the normal force exerted on M1 by the table.
Since M1 is on a table, the normal force F_normal_M1 is equal to the weight of M1, so F_normal_M1 = F_weight_M1.
For Mass 2 (M2):
The weight of M2 is given by F_weight_M2 = M2 * g.
The frictional force acting on M2 is given by F_friction_M2 = μs * F_normal_M2, where μs is the coefficient of static friction and F_normal_M2 is the normal force exerted on M2 by M1.
Since M2 is being pulled by the string connected to M1, the normal force F_normal_M2 is equal to the tension in the string, so F_normal_M2 = Tension.
To find the acceleration of each block, we need to consider the net force acting on each block.
For Mass 1 (M1), the net force F_net_M1 = Tension - F_friction_M1.
For Mass 2 (M2), the net force F_net_M2 = F_friction_M2 - F_weight_M2.
Since Mass 1 and Mass 2 are connected by a string, their accelerations will be equal, so we can set up the following equation:
F_net_M1 = M1 * a = Tension - F_friction_M1
F_net_M2 = M2 * a = F_friction_M2 - F_weight_M2
Now, let's substitute the equations for F_friction_M1, F_friction_M2, F_weight_M1, and F_weight_M2:
M1 * a = Tension - μs * F_normal_M1
M2 * a = μs * F_normal_M2 - F_weight_M2
Since F_normal_M1 = F_weight_M1 and F_normal_M2 = Tension, we can rewrite the equations as:
M1 * a = Tension - μs * (M1 * g)
M2 * a = μs * Tension - (M2 * g)
Now, solve these equations simultaneously to find the values of a (acceleration) and Tension.
b) To find the tension in the connecting string, we can use the equation:
M2 * a = μs * Tension - (M2 * g)
Rearranging the equation, we get:
Tension = (M2 * a + M2 * g) / μs
Substitute the values of M2, a, and μs into the equation to calculate the tension in the connecting string.