The diagram shows a block of mass m = 2.50 kg resting on a plane inclined at an

Question

The diagram shows a block of mass m = 2.50 kg resting on a plane inclined at an angle of theta = 3 degree to the horizontal. The coefficient of static friction between the block and the plane is mu static= 0.135, and the block is stationary but just on the point of sliding up the slope. The diagram shows the four forces acting on the block: an applied force F1 acting up the slope, the block's weight mg, the normal reaction force N and the force of static friction, Ff. In this case, the force of static friction acts down the slope, opposing the tendency of the block to move up the slope. Find the the maximum magnitude of the applied force F1 that can be exerted if the block is to remain stationary. Specify your answer by entering a number into the empty box below.

Explanation / Answer

Since the motion of a body is possible only along the incline plane So in calculating the maximum applied force we will consider only forces acting along the incline plane

here resolving the weight into two components

one component is along incline
and another component is along the perpendicular to the incline

Since the block has to remain stationary

net force along the incline must be zero

Fnet = F1-Ff-(mgsin(30)) = 0

but frictional force Ff = mu_s*m*g*cos(30) = 0.135*2.5*9.8*cos(30) = 2.86 N

m*g*sin(30) = 2.5*9.81*0.5 = 12.2625 N

Then required maximum applied force F1 = Ff+(m*g*sin(30)= 2.86+12.2625

F1 = 15.1225 N

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