First slide

Kinetic friction

Question

Two masses A and B of 10 kg and 5 kg respectively are connected with a string passing over a frictionless pulley fixed at the corner of a table as shown in fig. The coefficient of friction of A
with the table is 0.2 The minimum mass of C that may be placed on A to prevent it from moving is equal to

Moderate

A

15 kg
B

10 kg
C

5 kg
D

0 kg

Solution

Let T = tension in the string, f = frictional force between block A and table, x = minimum mass of C. For the system to be in equilibrium
f = T (for block A) ...(1)
where $f = μmg = μ (10 + x) g = 0 \cdot 2 (10 + x) g$
$∴ T = 0 \cdot 2 (10 + x) g . . . (2)$
For the equilibrium of B,
t = 5g …(3)
From eqs. (2) and (3), we get
$0 \cdot 2 (10 + x) g = 5 g or 2 + 0 \cdot 2 x = 5 or 0 \cdot 2 x = 3$
$or X = \frac{3}{0 \cdot 2} = 15 kg$ .

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