Answer:
Hey shaikaadil700 !
<u> </u><u>Lubricating</u><u> </u> of rough surfaces reduces friction.
Explanation:
• Lubricating is the smoothening or polishing of the surfaces

Answer:
Explanation:
Applied force, F = 18 N
Coefficient of static friction, μs = 0.4
Coefficient of kinetic friction, μs = 0.3
θ = 27°
Let N be the normal reaction of the wall acting on the block and m be the mass of block.
Resolve the components of force F.
As the block is in the horizontal equilibrium, so
F Cos 27° = N
N = 18 Cos 27° = 16.04 N
As the block does not slide so it means that the syatic friction force acting on the block balances the downwards forces acting on the block .
The force of static friction is μs x N = 0.4 x 16.04 = 6.42 N .... (1)
The vertically downward force acting on the block is mg - F Sin 27°
= mg - 18 Sin 27° = mg - 8.172 ... (2)
Now by equating the forces from equation (1) and (2), we get
mg - 8.172 = 6.42
mg = 14.592
m x 9.8 = 14.592
m = 1.49 kg
Thus, the mass of block is 1.5 kg.
Answer:
Fx = 4.92 [N]
Fy = 0.868 [N]
Explanation:
Let's take the 10 degrees as a measure from the horizontal component to the vector.
Thus taking the components in the X & y axes respectively:
Fx = 5*cos(10) = 4.92 [N]
Fy = 5*sin(10) = 0.868 [N]
<span>AS T1,T2,T3 are the tensions in the ropes,assuming that there are Three blocks of mass 3m, 2m, and m.T3 is the string between 3m and 2m,T2 is the string between 2m and m ,T1 is the string attached to m thus T1 pulls the whole set of blocks along, so it must be the largest. T2 pulls the last
two masses, but T3 only pulls the last mass, so T3 < T2 < T1.</span>