How Much Weight Must Block C Have to Prevent Block A from Sliding?

[djester555]

blocks A and B have weights of 45 N and 25 N, respectively.



block A is sitting on a table and block B is attached to block A by a piece of string on a frictionless pulley and block B is hanging off the table. also block C is on top of block A



Figure 6-48
(a) Determine the minimum weight of block C to keep A from sliding if µstat between A and the table is 0.20.

(b) Block C suddenly is lifted off A. What is the acceleration of block A if µkin between A and the table is 0.15?
2 m/s2




A)
block a = 45n
block b = 25n
block c = 45 -25
block C = 20N i don't think this is right

B) can't solve without part A

 

[rl.bhat]

Find out the forces acting on A and C. Similarly find out the force acting on B. Find the acceleration of the system. Find the condition for the equilibrium.

 

[tomcue]

I would approach this problem by first identifying the different forces acting on each block. For block A, there is the weight force pulling it downward and the normal force from the table pushing it upward. For block B, there is the weight force pulling it downward, the tension force from the string pulling it upward, and the normal force from block A pushing it upward. For block C, there is only the weight force pulling it downward.

For part A, we need to determine the minimum weight of block C to keep block A from sliding. This means that the tension force from the string must be equal to or greater than the weight force of block A. We can set up an equation: T ≥ mg, where T is the tension force, m is the mass of block C, and g is the acceleration due to gravity. We can also substitute the weight of block A for mg, giving us T ≥ 45 N. Since the coefficient of static friction (µstat) between block A and the table is given as 0.20, we can use the equation T = µstatN, where N is the normal force from the table. Substituting in our known values, we get µstatN ≥ 45 N. Solving for N, we get N ≥ 225 N. This means that the normal force from the table must be at least 225 N to prevent block A from sliding. Therefore, the minimum weight of block C would be 225 N - 45 N = 180 N.

For part B, we are given that block C is suddenly lifted off block A. This means that the tension force from the string is no longer present, and the only forces acting on block A are its weight force and the normal force from the table. We can use the equation Fnet = ma, where Fnet is the net force on block A, m is its mass, and a is its acceleration. Substituting in the known values, we get Fnet = 45 N - µkinN, where µkin is the coefficient of kinetic friction between block A and the table. We are given that µkin = 0.15, so we can solve for N: N = (45 N - Fnet)/0.15. Since we are looking for the acceleration of block A, we can set Fnet = ma and solve for a: a = (45 N - ma)/0.15.