Free body diagram Incline Problem

[imdapolak]

Problem: Block 1 is 7kg and is resting on a table with friction. Block 2.5kg is connected by a massless cord and is hanging over the side of the table. ( assume there is no friction on the cord where it goes over the side of the table.) The two masses are stationary but on the verge of sliding.
a.) Draw a free body diagram
b.) What is the magnitude of the frictional force holding the blocks in place? (static friction.)
c.) What is the coefficient of static friction between the table and the block?
d.) If you give the blocks a slight nudge, they begin to move. If the coefficient of kinetic friction is 0.54, what is the frictional force (kinetic) now that they are moving?

I draw the free body diagram and understand their is Fg, Fnormal, Ffriction, Ftension on block 1, then block 2 has Ftension and Fg. Not really sure what to do next to solve for frictional force. By the way what's the difference between static and kinetic friction? My book eludes this information to me.

 

[Kurdt]

There is obviously a force from the block hanging over the table trying to pull the one on the table off. If they're on the brink of moving then the frictional force must be equal to the force trying to move them. Here is a link that has a couple of articles on the differences between the frictions.

http://hyperphysics.phy-astr.gsu.edu/hbasees/frict2.html

 

[tomcue]

a.) The free body diagram for this problem would include the following forces: Fg (gravitational force) acting downwards on both blocks, Fnormal (normal force) acting upwards on block 1, Ffriction (frictional force) acting to the left on block 1, and Ftension (tension force) acting upwards on block 2.

b.) The magnitude of the frictional force holding the blocks in place would be equal to the product of the coefficient of static friction and the normal force (Ffriction = μs * Fnormal). However, since the blocks are on the verge of sliding, the frictional force would be equal to the maximum possible value, which is the product of the coefficient of static friction and the normal force (Ffriction = μs * Fnormal).

c.) The coefficient of static friction between the table and the block can be calculated by dividing the maximum frictional force (Ffriction) by the normal force (Fnormal). So, μs = Ffriction / Fnormal.

d.) Once the blocks start moving, the frictional force would change from static to kinetic. The coefficient of kinetic friction (μk) would be different from the coefficient of static friction (μs). The frictional force (Ffriction) in this case would be equal to the product of the coefficient of kinetic friction and the normal force (Ffriction = μk * Fnormal). So, to calculate the frictional force when the blocks are moving, we would need to know the value of μk and Fnormal.