Question

Block A, block E and load C are connected using cables as shown in Figure 4 and mass of each 20kg, 5kg and 5kg respectively. The inclined plane is smooth, and the coefficient of kinetic friction between the horizontal surface and block E is µk = 0.25. Angle of inclined plan is 600. i. ii. iii. iv. v. Draw a free body diagram (FBD) of the Block A, block E and load C. Determine kinetic friction force applied on the block E. Calculate acceleration of the system. Determine tension developed in the each cable. Determine resultant force applied on the block A parallel to the inclined plane.

Answer 1

The FBD of he system is given below,

Kinetic friction force on the block E$=\mu_k N$

$=4\times 9.8\times 0.25 N$

$=9.8N$

Let the acceleration of the whole system is a,

For block C, $5g-T_2= 5a ...(i)$

For block A, $T_2-T_1+20g\sin60^\circ=20a...(ii)$

For the block E,$T_1-f_s=4a...(iii)$

From equation (i), (ii) and (iii)

$\Rightarrow5g+20g\sin 60^\circ-9.8=29a$

$\Rightarrow a = \frac{208.94}{29}m/s^2$

$\Rightarrow a =7.2m/s^2$

Substituting the value of a in the equation (i)

$T_2=5g-5a = (49-36)N$

$\Rightarrow T_2 =13N$

$T_1=4a+f_s =4\times 7.2+9.8 =28.8+9.8=38.6N$

$T_1=38.6N$

Resultant force on the block A,

$\Rightarrow F= 20g\sin60^\circ +T_2-T_1$

$\Rightarrow F= 20\times 9.8 \sin 60^\circ +13-38.6$

$\Rightarrow F = 144.14N$