Question

A) A 6.0 kg mass sits atop a larger 9.0 kg mass which itself sits on a horizontal table top. A third, 3.0 kg mass hangs vertically just barely touching the right face of the large mass. A massless inflexible string connects the hanging mass to the top mass passing over a frictionless pulley (P). All horizontal surfaces have coefficients of static and kinetic friction of 0.80 and 0.40 respectively. All vertical surfaces are frictionless.

There is a little tap directly down on the hanging mass, just enough to cause this
block to slide. Assume this motion applies to parts 3, 4 and 5.

3) The 9.0 kg mass does not move. What is the acceleration of m2; tension now
in the string?
4) What is the magnitude of the net horizontal force on the pulley? What is the
magnitude of net force on the support rod (S) by the pulley?
5) What is the magnitude and direction of the total friction force on the large
mass, if any, at the interface between the table top and this mass?

Answer 1

Answer on Question #74798, Physics / Mechanics | Relativity

A) A 6.0 kg mass sits atop a larger 9.0 kg mass which itself sits on a horizontal table top. A third, 3.0 kg mass hangs vertically just barely touching the right face of the large mass. A massless inflexible string connects the hanging mass to the top mass passing over a frictionless pulley (P). All horizontal surfaces have coefficients of static and kinetic friction of 0.80 and 0.40 respectively. All vertical surfaces are frictionless.

There is a little tap directly down on the hanging mass, just enough to cause this block to slide. Assume this motion applies to parts 3, 4 and 5.

3) The 9.0 kg mass does not move. What is the acceleration of m2; tension now in the string?

Solution.

$$\underline{F}_{fr} = k \cdot P = k \cdot mg$$ $$F_{fr.m2} = 0,4 \cdot 6 \cdot 9,81 = 23,54 \, N$$ $$F_{ten} = m \cdot g$$ $$F_{ten} = 3 \cdot 9,81 = 29,43 \, N$$ $$F = F_{ten} - F_{fr.m2} = 29,43 - 23,54 = 5,89 \, N$$ $$F = m \cdot a$$ $$a = \frac{F}{m} = \frac{5,89}{6} = 0,98 \, \frac{m}{s^2}$$

Answer: $F_{ten} = 29,43 \, N$, $a = 0,98 \, \frac{m}{s^2}$

4) What is the magnitude of the net horizontal force on the pulley? What is the magnitude of net force on the support rod (S) by the pulley?

$$F_{\text{net horizontal}} = F_{ten} - F_{fr.m2} = 29,43 - 23,54 = 5,89 \, N$$ $$F_{\text{net force on the support rod (S) by the pulley}} = 2 \cdot F_{ten} \cdot \sin \frac{\alpha}{2} = 2 \cdot 29,43 \cdot \sin \frac{90}{2} = 2 \cdot 29,43 \cdot 0,7 =$$ $$= 41,2 \, N$$

Answer: $F_{\text{net horizontal}} = 5,89 \, N$; $F_{\text{net force on the support rod (S) by the pulley}} = 41,2 \, N$

5) What is the magnitude and direction of the total friction force on the large mass, if any, at the interface between the table top and this mass?

**Answer:** Total friction force on the large mass equals to zero.

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