A block of mass m = 4.50 kg rides on top of a second block of mass M = 12.0 kg.

Question

A block of mass m = 4.50 kg rides on top of a second block of mass M = 12.0 kg. A person attaches a string to the bottom block and pulls the system horizontally across a frictionless surface as in figure (a). Friction between the two blocks keeps the 4.50 kg block from slipping off. If the coefficient of static friction is 0.380, answer the following.
(a) What maximum force can be exerted on the string without causing the 4.50 kg block to slip?
Correct: Your answer is correct. 61.45 N

(b) Use the system approach to calculate the acceleration.
Correct: Your answer is correct. 3.724 m/s2

EXERCISE HINTS:
Use the values from PRACTICE IT to help you work this exercise. Suppose instead that the string is attached to the top block as shown in figure (b). Find the maximum force that can be exerted by the string on the block without causing the top block to slip.
N NEED THIS LAST PART.....I haven't been able to figure it out.

Explanation / Answer

A) let the maximum force exerted is F= (4.5+12)a from the FBD of block A acceleration of block is also a thus force = 4.5*a this force is resisted by friction 4.5*a = .380*4.5*9.8 a = 3.724 m/ sec2 thus MAX force on string = (4.5+12)*3.724 = 61.446 For last part A) let the maximum force exerted is F= (4.5+12)a from the FBD of block A acceleration of block is also a thus force = 4.5*a this force is resisted by friction 4.5*a = -.380*4.5*9.8 + (4.5+12)a a = 1.3965 m/ sec2 thus MAX force on string = (4.5+12)*1.3965 =23.04

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