2 blocks are in contact on a horizontal table. A force is applied on the first block at a 30* angle. The blocks move to the right side.
If m1 = 2kg and
m2 = 1kg and
F = 6 N
kinetic friction at the middle of the 2 blocks and the table = 0.250
Find the "contact force" between the 2 blocks.
You need to specify if the 30 degree angle is up or down from horizontal. That will affect the friction force.
it is down from horizontal drwls,
Thank you for your time
Forward applied force = F cos 30 = 5.196 N
Friction force on m1 = (m1*g+Fsin30)*0.25 = 5.65 N
Friction force on m2 = (m2*g*0.25)
= 2.45 N
Acceleration of m1 and m2:
a = (5.65+2.45)/(m1 + m2) = 2.7 m/s^2
The force between the blocks, F', minus friction on m2, is what accelerates m2.
F' - 2.45 = m2*a
Solve for F'
the angle is at the top left corner of the first block
It should not matter where it was applied, as long as it was on the first block, in the downward direction you said.
ok thank you drwls!
To find the contact force between the two blocks, we need to consider the forces acting on each block individually.
Let's start by drawing a free-body diagram for each block:
For the first block (m1):
1. The force applied at a 30° angle (F) pushes the first block to the right.
2. There is a contact force (Fc) acting on the left side of the first block due to the second block.
3. There is a normal force (N1) acting on the first block in an upward direction.
4. There is kinetic friction (fk1) acting in the opposite direction of the motion.
For the second block (m2):
1. There is a contact force (Fc) acting on the right side of the second block due to the first block.
2. There is a normal force (N2) acting on the second block in an upward direction.
3. There is kinetic friction (fk2) acting in the opposite direction of the motion.
Now, let's calculate the force of kinetic friction acting on each block:
fk1 = μ * N1
fk2 = μ * N2
Given that the coefficient of kinetic friction (μ) is 0.250, we need to compute the normal forces for each block:
N1 = m1 * g
N2 = m2 * g
where g is the acceleration due to gravity, approximately 9.8 m/s^2.
N1 = 2 kg * 9.8 m/s^2
N2 = 1 kg * 9.8 m/s^2
Next, we can calculate the contact forces on each block:
Fc = F * cos(30°)
Finally, to find the contact force between the two blocks, we need to make sure the force exerted by block 1 on block 2 is equal to the force exerted by block 2 on block 1. Therefore:
Fc = Fc
Substituting the expressions for Fc, we can solve for the unknown contact force.