Answer to Question #92363 in Mechanics | Relativity for Fatima

Question

1. The force applied to the. System given in figure. 1 has a magnitude of. 50N. The blocks have. Masses M1 = 1kg and M2 = 2kg. The coefficient of kinetic friction between each block and surface is 0.150. Hist: Assume that g=9.8m/s2.
A) Draw a free body diagram for each block
B)Find the acceleration of the system.

Answer 1

Define N₁ = force of support reaction acting on M1, N₂ = force of support reaction acting on M1,

T = tension force, F_f1 = friction force acting on M₁, F_f2 = friction force acting on M₂, k = coefficient of kinetic friction, F_T = sum of forces acting on the system of M₁ and M₂.

a)

b)

Since T is internal force of the system and

$$\overrightarrow{N_1}+M_1\overrightarrow{g}=0, \overrightarrow{N_2}+M_2\overrightarrow{g}=0$$

$$\overrightarrow{F_T}=\overrightarrow{F}-\overrightarrow{F_{f1}}-\overrightarrow{F_{f2}}$$

$$|\overrightarrow{F_T}|=|\overrightarrow{F}|-|\overrightarrow{F_{f1}}|-|\overrightarrow{F_{f2}}|$$

$$|\overrightarrow{F_T}|=|\overrightarrow{F}|-kM_1g-kM_2g$$

Using Newton's Second Law acceleration of blocks

$$a=\frac{|\overrightarrow{F_T}|}{M_1+M_2}=\frac{|\overrightarrow{F}|-(kM_1g+kM_2g)}{M_1+M_2}=$$

$$\frac{|\overrightarrow{F}|-kg(M_1+M_2)}{M_1+M_2}=\frac{|\overrightarrow{F}|}{M_1+M_2}-kg=$$

$$\frac{50N}{1kg+2kg}-0.15\cdot9.8m/s^2\approx15.19667m/s^2$$

Answer: a = 15.19667m/s².