Figure 6-14 shows a block of mass in connected to a block of mass M 2.00 kg, bot

Question

Figure 6-14 shows a block of mass in connected to a block of mass M 2.00 kg, both on 45 degree inclined planes where the coefficient of static friction is 0.28. Find the (a) minimum and (b) maximum values of m for which the system is at rest. A 2.5 kg block is initially at rest on a horizontal surface. A horizontal force F rightarrow of magnitude 6.0 N and a vertical force P rightarrow are then applied to the block (Fig. 6-15). The coefficients of friction for the block and surface are mu_s = 0.40 and M_k = 0.25. Determine the magnitude of the frictional force acting on the block if the magnitude of P rightarrow is (a) 8.0 N, (b) 10 N, and (c) 12 N.

Explanation / Answer

Writing equations of motion

mg*sin(45) + fss = tension T

Tension T +fs2 = M*g*sin(45)

m*g*sin(45)+fs1+fs2 = M*g*sin(45)

m*g*sin(45) = (2*9.8*sin(45)) - (0.28*9.8*cos(45)(m+2))

m*g*sin(45) = 13.85 - 1.94*(m+2)

8.86*m = 9.97

m = 9.97/8.86 = 1.12 kg

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frictional force = f = mu*N

for a) P = 8N

then f = 0.4*8 = 3.2 N

3.2 N < 6 N

so the bodt doesn't move

then the static frictional force comes into play

required frictional force is fs = mu_s*P = 0.4*8 = 3.2

if P = 10 N

then frictional force is fs = mu_s*P = 0.4*10 = 4 N

if P = 12 N

frictional force is fs = mu_s*P = 12*0.4 = 4.8 N

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