Block against cart and stacked blocks with friction

[midnightassassinx]

1) A block is placed agains the vertical front of a cart. What acceleration must the cart have in order for the block to not fall? The coefficient of static friction between the block and the cart is Us (mu s). How would an observer on the cart describe the behavior of the block? (Solve with variables)

2) A block B, with a mass Mb, rest on block A, iwth a mass of Ma, which in turn is on a horizontal tabletop. The coefficient of kinetic friction between block A and the tabletop is Uk (mu k), and the coefficient of static friction between block A and block B is Us (mu s). A massless string is attached to block A and passes over a frictionless, massless pulley and block C is suspended from the other end of the string. What is the largest mass Mc that block C can have so that blocks A and B still slide together when the system is released from rest.

 

[Sirus]

Couple things. First, please always remember to show us what you have done in attempting a problem and where you got stuck. Second, always begin these types of problems with a free-body diagram (or several, as in question 2). Then we can help you.

 

[wais]

1) To find the acceleration of the cart, we can use Newton's second law, which states that the net force on an object is equal to its mass multiplied by its acceleration (F=ma). In this case, the net force on the block is the force of gravity pulling it down (mg) and the force of friction pushing it up (Fs). The observer on the cart would see the block remain stationary, so the acceleration of the block must be zero. Therefore, the net force on the block must also be zero, meaning that the force of gravity is equal and opposite to the force of friction (mg=Fs). We can substitute the formula for friction (Fs=mu s*N) and the formula for weight (mg=mg) to get mu s*N=mg. We can then solve for the normal force (N) by dividing both sides by mu s, giving us N=mg/mu s. The normal force is equal to the force of friction, so we can substitute this into the formula for friction to get Fs=mu s*(mg/mu s). The mu s terms cancel out, leaving us with Fs=mg. We can then substitute this into the formula for net force to get mg=ma. Finally, we can solve for acceleration by dividing both sides by the mass of the cart (m), giving us a=g. Therefore, the cart must have an acceleration of g (9.8 m/s^2) for the block to not fall.

2) To find the largest mass that block C can have, we can use the concept of equilibrium, which states that the net force and net torque on an object must be equal to zero for it to be in equilibrium. In this case, we can consider the system of blocks A and B as one object, and block C as a separate object. For the system of A and B to remain in equilibrium, the net force on them must be zero, meaning that the force of friction (Fs) must be equal to the force of weight (mg) pulling them down. We can use the formula for friction (Fs=mu s*N) and the formula for weight (mg=mg) to get mu s*N=mg. We can then solve for the normal force (N) by dividing both sides by mu s, giving us N=mg/mu s. This is the maximum normal force that can be exerted on the system before it starts to slide.