two blocks a and b having masses 5 kg and 10 kg respectively connected
Answers
The minimum mass c is equal to 15 Kg.
Explanation:
Correct statement:
Two masses A and B of 10 kg and 5 kg, respectively , are connected with a string passing over a friction less pulley fixed at the corner of a table as shown. The coefficient of static friction between A and the table is 0.2. The minimum mass C that should be placed on A to prevent it from moving is equal to
Solution:
fL = μN
fL = μ(10 + mc)g
fL = T = 5 g
fL = 5 g
μ(10 + mc)/g = 5g
10 + mc = 5 / μ = 5/2 x 10
mc = 50 / 2 - 10
mc = 15 Kg
Thus the minimum mass c is equal to 15 Kg.
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Two blocks A and B of mass 10 kg and 40 kg are connected by an ideal string as shown in the figure neglect the masses of pulleys and effect of friction in the pulley between the blocks and the inclines.(wedge is fixed) ?
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