I got two questions
Three blocks on a fricitonless horizontal surface are in contact with each other
a force F is applied to block 1 mass m1
Draw a free body diagram for each block
ok done
the accelration of the system (in terms of m1, m2, and m3)
ok done I got
a = (m1 + m2 + m3)^-1 (force F)
the force of contact that each block exerts on its neighbor
ok no idea how to do this
----->Force F | m1 |--->contact force | m2 |---->contact force | m3 |
ok so how do i do this
ok and
a block (mass m1) on a smooth horizontal surface, connected by a thin cord that passes over a pulley to a second block (m2), which hangs vertically.
Draw a free body diagram for each block showing the force of gravity on each, the force (tension) exerted by the cord, and any normal force
ok done
determine formulas for the acceleration of the system and for the tnesion in the cord. Ignore friction and the masses of the pulley and cord
ok well isn't the force of tension just the opposite in direction but equal in magntiude of the force of gravity on second hanging block?
Also isn't the accleeration zero???
Isn't the net force zero???
ok i do not know how to find the acceleration of the system
Ok I guess I'm drawing the free body diagram wrong how do I draw it because this is not making much sense to me
Why did you start this question by re-asking the three blocks question that we have already explained to you? I assume that you followed directions and are through with that one.
As for the problem with the two masses, cord and pulley: the acceleration is NOT zero in a frictionless situation. The dangling mass m2 exerts a weight force on the cord, which allows both masses to move with the same acceleration, a. What equations of motion have you derived from your FBDs?
The tension force T in the cord does not equal the weight force W = m2*g when the masses are accelerating.
You should have two equations in two unknowns: T and a.
T = m1*a
m2*g - T = m2*a
m2*g = (m1 + m2) a
Solve for a and T
is this right?
------Force F onto m1--->|m1| ---contact force onto m2--->|m2|-----contact force onto m3---->|m3|
No, it is not right. Force and mass have different dimensions.
Mass has to be multiplied by an acceleration rate to get force.
To find the acceleration of the system in the first scenario, where three blocks are in contact, you can use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration.
Start by drawing a free body diagram for each block.
For block 1:
- Mark an arrow pointing to the right to represent the applied force F.
- Label the mass of block 1 as m1.
- Since the surface is frictionless, there are no other forces acting on block 1.
For block 2:
- Since block 1 and block 2 are in contact, there is a contact force between them. Draw an arrow pointing to the right to represent this force.
- Label the mass of block 2 as m2.
- There is no applied force acting on block 2, and since the surface is frictionless, there are no other forces.
For block 3:
- Since block 2 and block 3 are in contact, there is a contact force between them. Draw an arrow pointing to the right to represent this force.
- Label the mass of block 3 as m3.
- There is no applied force acting on block 3, and since the surface is frictionless, there are no other forces.
Now, to calculate the acceleration of the system, you need to consider the forces acting on block 1. The net force on block 1 is equal to the applied force F, and according to Newton's second law, it is also equal to the mass of block 1 multiplied by its acceleration.
Therefore, we have:
F = m1 * a
To find the overall acceleration of the system, we can assume that the contact forces between the blocks are equal in magnitude. So the total force acting on block 1 can be calculated as the sum of the contact forces with block 2 and block 3.
Therefore, we have:
F = (force of contact between block 1 and block 2) + (force of contact between block 1 and block 3)
Since the forces are equal between the blocks, we can rewrite this as:
F = 2 * (force of contact)
Now, substitute this into the previous equation:
2 * (force of contact) = m1 * a
Rearranging the equation, you get:
force of contact = (m1 * a) / 2
Since force of contact is the same for both block 1 and block 2, we can write:
force of contact = ((m1 + m2) * a) / 2
Similarly, the force of contact between block 2 and block 3 is also ((m2 + m3) * a) / 2.
So, in conclusion:
- The acceleration of the system is given by a = (m1 + m2 + m3)^-1 * F.
- The force of contact that each block exerts on its neighbor is ((mass of block 1 + mass of block 2) * acceleration) / 2 and ((mass of block 2 + mass of block 3) * acceleration) / 2.
Now, moving on to the second scenario, where two blocks are connected by a cord passing over a pulley:
Draw a free body diagram for each block:
For block 1:
- Mark an arrow pointing to the right to represent the force of tension in the cord.
- Label the mass of block 1 as m1.
- Mark an arrow pointing downward to represent the force of gravity acting on block 1.
- Since the surface is smooth, there is no friction, and thus no other forces acting on block 1.
For block 2:
- Mark an arrow pointing downward to represent the force of gravity acting on block 2.
- Label the mass of block 2 as m2.
- Since the cord is connected to block 2, mark an arrow pointing to the left to represent the force of tension.
- There are no other forces acting on block 2.
Regarding your questions about the tension, acceleration, and net force:
- The force of tension in the cord is indeed equal in magnitude but opposite in direction to the force of gravity acting on block 2.
- The acceleration of the system is not necessarily zero. It depends on the masses of the two blocks and the forces involved. If the masses are unequal, the system will experience an acceleration.
- The net force on the system is not zero. The net force is equal to the force of tension minus the force of gravity. If the masses are unequal, the net force will be nonzero, resulting in acceleration.
To calculate the acceleration of the system, you can start by considering the forces acting on block 1. The net force acting on block 1 is equal to the force of tension minus the force of gravity acting on block 1, and according to Newton's second law, it is also equal to the mass of block 1 multiplied by its acceleration.
Therefore, we have:
Force of tension - Force of gravity on block 1 = m1 * a
The force of tension is equal in magnitude to the force of gravity on block 2, which is equal to m2 * g (where g is the acceleration due to gravity).
Substituting this into the equation, you get:
Force of tension - m2 * g = m1 * a
Now, rearranging the equation, you can solve for the acceleration:
a = (Force of tension - m2 * g) / m1
Additionally, since the force of tension in the cord is equal to the force of gravity on block 2, you can substitute in the equation:
a = (m2 * g - m2 * g) / m1
Simplifying further, you get:
a = 0
This means that the acceleration of the system is indeed zero in this scenario.
Finally, to find the formula for the tension in the cord, you can substitute the acceleration (which is zero) into the previous equation:
Force of tension = m2 * g
Therefore, the tension in the cord is equal to the force of gravity on block 2, which is m2 * g.