Question
Block B, with mass 5.50kg, rests on block A, with mass 7.50 kg, which in turn is on a horizontaltabletop. There is no friction between block A and thetabletop, but the coefficient of static friction between blockA and block B is 0.78. A light string attached to block A passesover a frictionless, massless pulley, and block C issuspended from the other end of the string. What is the largestmass that block C can have so that blocks A andB still slide together when the system is released fromrest?
kg
Explanation / Answer
The coefficient of static friction can give you the forceneeded to overcome the static friction by this equation. Force required(fs)=s(coeffiecint)*n(normal). The normal of B is 5.50*9.8=53.9 N. The force that block Asupports. So teh force required to covercome the friction is fs=.78*53.9=42.042 N. This is in the direction that C ispulling. Since block B resists the change of motion and is not attachedto the string. The maximum force that can be applied to block A is42.042. This means that the max acceleration is42.042/(5.5+7.5)=3.234 m/s^2. Now we need to find the mass of block C that produces thatacceleration. The system accelerates like this. The blocks are beign pulled down by gravity 5.5+7.5=13 13*9.8=127.4 N is the normal for block A and B For block C. We know that the total accelerationof blockA and B needs to be 3.234 m/s^2 so that is how much block Cneeds. mC*g=(mA+mB+mC)*a We want to find mC. 9.8mC=42.042+3.234*mC mC=6.4 kg. Which is your final answer.