Inclined Plane

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A block of mass 10 kg rests on an inclined plane making an angle of 30° with the horizontal. If the coefficient of static friction is 0.5, what is the minimum force required to move the block up the plane? (g = 10 m/s²)

RankGuru · Accepted Answer

1. **Force Analysis**: To move the block **up**, the applied force $P$ must overcome both gravity's component and maximum friction.
   - Force down the plane: $F_g = mg \sin 	heta = 100 	imes 0.5 = 50$ N.
   - Normal force: $N = mg \cos 	heta = 100 	imes 0.866 = 86.6$ N.
2. **Friction Calculation**:
   - $f_{max} = \mu_s N = 0.5 	imes 86.6 = 43.3$ N.
3. **Summation**:
   - $P_{min} = F_g + f_{max} = 50 + 43.3 = 93.3$ N.
4. **Result**: A force of **93.3 N** is required to break static equilibrium and initiate motion upward.

Inclined Plane

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