2 blocks are connected by a rope of negligible mass are being dragged by a horizontal force. Suppose F= 68.0N, m1= 12.0kg, m2= 18.0kg and the coefficient of kinetic friction between each block and the surface is 0.100.
a.) draw free body diagram for each block
b.) determine the tension T and the magnitude of the acceleration of the system.
Ffk= uk * Fn (Ffk=force of kinetic friction) (uk= coefficient of friction) (Fn= normal force)
F-T= m2*a
T-f= m1a or is it T= m1*a (I changed the equation that I saw for a system with objects being pulled but they said it was T= m1*a however I'm thinking that this doesn't include friction into it since the one with the first box connected directly to the rope being pulled does take into account the tension force and it subtracts it from the force that pulls the box ...so I was thinking maybe they didn't include the friction since it would also be a opposing force to the tension..am I correct?)
a= F/m1 + m2
I really really really really need help in how i incorperate the coefficient of friction into this problem..I was thinking that that that would be the force it is pulled with but I really don't know what to do..sinc they give you a force that the boxes are pulled with
Can someone give me a push in the right direction...especially how to relate the forces to the coefficient of friction..
THANKS
I cannot draw the F.B.D for you here, but assume that m1 is the mass being pulled by force F, and that m2 is the mass that is trailing behind, pulled by rope tension T. T also ags as a reverse force on mass m1.
The equations of motion are
F - Ff1 - T = m1*a
T - Ff2 = m2*a
Adding the two equations together cancels out the t and lets you solve for a.
F - Ff1 - Ff2 = (m1 + m2)*a
a = (F - Ff1 - Ff2)/(m1 + m2)
You already have written the correct equation for the Ffk friction terms.
Once you a from the above equation, you can solve for T.
T = Ff2 + m2*(F - Ff1 - Ff2)/(m1 + m2)
THANKS SO MUCH DrWls !! =D
a.) To draw the free body diagrams, we need to consider the forces acting on each block.
For block 1 (m1 = 12.0 kg):
1. The weight of the block, directed downwards (mg).
2. The tension force T, directed to the right.
3. The force of kinetic friction (Ffk1) between block 1 and the surface, directed to the left.
For block 2 (m2 = 18.0 kg):
1. The weight of the block, directed downwards (mg).
2. The tension force T, directed to the left.
3. The force of kinetic friction (Ffk2) between block 2 and the surface, directed to the right.
b.) To determine the tension force T and the magnitude of the acceleration of the system, we need to consider the forces acting on both blocks and the relationship between their accelerations.
For block 1 (m1 = 12.0 kg):
Net force in the horizontal direction = F - Ffk1 - T
F = 68.0 N (given)
Ffk1 = uk * Fn, where uk = 0.100 (given)
Fn = m1 * g, where g = 9.8 m/s^2 (acceleration due to gravity)
Using the equation F - Ffk1 - T = m1 * a (where a is the acceleration of block 1), we can solve for T.
For block 2 (m2 = 18.0 kg):
Net force in the horizontal direction = T - Ffk2
Ffk2 = uk * Fn, where uk = 0.100 (given)
Fn = m2 * g, where g = 9.8 m/s^2 (acceleration due to gravity)
Using the equation T - Ffk2 = m2 * a (where a is the acceleration of block 2), we can solve for a.
The magnitude of the acceleration of the system can also be calculated using the equation a = (F - Ffk1 - T) / (m1 + m2).
Let's substitute the known values into the equations and solve for T and a.
To incorporate the coefficient of friction into this problem, you need to consider the force of friction acting on each block. Here's how you can proceed:
a.) Free body diagram for each block:
For the first block (m1):
- Draw an arrow representing the weight (mg) downward.
- Draw an arrow representing the normal force (Fn) upward (since the block is on a horizontal surface).
- Draw an arrow representing the force of kinetic friction (Ffr) in the opposite direction of motion.
For the second block (m2):
- Draw an arrow representing the weight (mg) downward.
- Draw an arrow representing the normal force (Fn) upward.
- Draw an arrow representing the force of kinetic friction (Ffr) in the opposite direction of motion.
b.) Determining the tension (T) and the acceleration (a) of the system:
- Start with the equation for the second block: F - T - Ffr = m2 * a. Since the blocks are connected by a rope, the tension experienced by the first block is also the tension experienced by the second block.
- Next, consider the first block: T - Ffr = m1 * a.
To relate the coefficient of friction to the force of kinetic friction (Ffr), you can use the equation: Ffr = uk * Fn, where uk is the coefficient of friction and Fn is the normal force. Since the normal force on each block is equal to the weight (Fn = mg), you can write: Ffr = uk * mg.
Now, you can substitute the expressions for Ffr into the equations for each block:
- For the second block: F - T - uk * m2 * g = m2 * a.
- For the first block: T - uk * m1 * g = m1 * a.
To solve for T and a, you have two equations with two unknowns. You can solve these equations simultaneously to find the values of T and a.
Note: The equation T = m1 * a (which you mentioned) assumes there is no friction and is not applicable in this case since there is a coefficient of kinetic friction given.