Need freebody diagrams and please show work. The answer is 21.6N 4. A 4.0kg bloc

Question

Need freebody diagrams and please show work. The answer is 21.6N

4. A 4.0kg block rests on top of a 5.0kg block that rests on a frictionless table. The coefficient of friction between the two blocks is such that the blocks start to slip when the horizontal force F applied to the lower block is 27N. Suppose now that a horizontal force is only applied to the top block. What is the maximum value of this force so that the blocks slide without slipping relative to each other (25 points)? 4.0kg F-27N 5.0kg Ans 21.6N

Explanation / Answer

Assume the magnitude of the friction between the two blocks is f at the moment just before the blocks start to slip. According to Newton’s third law, the direction of the frictional force on the bottom block is pointing horizontally towards left and the frictional force on the top block is pointing horizontally towards right.

F - f = M1*a          ....................(1)

f = M2*a               ....................(2)

adding equation (1) and (2)

F = (M1 + M2)a

a = F / (M1 + M2)

a = 27 N / 9 kg

a = 3 m/s^2

put in eq (2)

f = 12 N

for the maximum force that can be applied to the top block without causing slipping between the two blocks,

F - f = M2*a'          ....................(3)

f = M1*a'               ....................(4)

equation (4) becomes,

a' = f / M1

a' = 12 N / 5 kg

a' = 2.4 m/s^2

equation (3) becomes,

F = f + M2*a'

F = 12 N + (4 kg x 2.4 m/s^2)

F = 21.6 N

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.