First slide

Kinetic friction

Question

A block of mass 1 kg is at rest on a horizontal table. The coefficient of static friction between the block and the table is 0.50. If g = 10 m/s², then the magnitude of a force acting upwards at an angle of 60° from the horizontal that will just start the block moving is

Difficult

A

5N
B

5.36 N
C

74.6 N
D

10 N

Solution

$R + Psin 60^{\circ} = mg or R = mg - Psin 60^{\circ}$
Now, frictional force,
$\begin{array}{rrlrlrlrlrl} f & = μR \\ ∴ & f & = μ [mg - Psin 60^{\circ}] . . . (1) \end{array}$
For the body to just move
$\begin{array}{l} Pcos 60^{\circ} = f = μ [mg - Psin 60^{\circ}] \\ ∴ \frac{P}{2} = 0 \cdot 5 [1 \times 10 - \frac{P \sqrt{2}}{2}] or P = 10 - \frac{P \sqrt{3}}{2} \end{array}$
$or P + P (\frac{\sqrt{3}}{2}) = 10 or P = 5 \cdot 36 N .$

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