As shown in figure force F = 100 N is applied horizontally on a block of mass 4 kg. Which is in contact with a wall. Such that it does not fall. The coefficient of friction between the block and the wall is ......... .
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Q. A force of 100N is applied on a block of mass 3 kg as shown in the figure. The coefficient of friction between the wall and block is 0.6. The magnitude of the force exerted by the wall on the block is?
Solution - The force is acting on the body in 30 degree perpendicular to the wall.
The force is acting on the body in 30 degree perpendicular to the wall.From the free body diagram we can resolve that,
The force is acting on the body in 30 degree perpendicular to the wall.From the free body diagram we can resolve that,100cos300+F=30N
The force is acting on the body in 30 degree perpendicular to the wall.From the free body diagram we can resolve that,100cos300+F=30NF=30−100×21=−20N
The force is acting on the body in 30 degree perpendicular to the wall.From the free body diagram we can resolve that,100cos300+F=30NF=30−100×21=−20NHence the magnitude of force exetred by the wall is 20N downward
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