First slide

Friction on horizontal surface

Question

A force F pushes a block weighing l0 kg against a vertical wall as shown in the Figure. The coefficient of friction between the block and wall is 0.5. The minimum value of F to start the upward motion of block is : (g = 10m/s²)

Moderate

A

500 newton
B

400 newton
C

250 newton
D

can't be determined

Solution

FBD of block
From diagram
$N = Fcos 37^{\circ} = F (\frac{4}{5}) = \frac{4 F}{5}$
To start upward motion
$\begin{array}{l} Fsin 37^{\circ} = 10 g + μN \\ Therefore, \frac{3}{5} F = 100 + 0 .5 (\frac{4}{5} F) \end{array}$

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