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I. Multiple Choice Questions · Q2

Q.An object of mass mm is held at rest against a vertical wall by pressing it with a horizontal force FF, as shown in the figure.

a block of mass m held against a vertical wall by a horizontal force F, with normal reaction, friction and weight marked — Class 12 Physics question
Figure
(IIT JEE 1994) The minimum value of the force FF is
(a) less than mgmg
(b) equal to mgmg
(c) greater than mgmg
(d) cannot be determined
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✓ Free question

Step 1. Two forces act horizontally/vertically on the block: the applied force FF (horizontal) and the wall's normal reaction N=FN=F; vertically, gravity mgmg acts down and static friction fs≤μN=μFf_s\le\mu N=\mu F acts up.

Step 2. For the block to remain at rest (not slide down), friction must supply the full weight: fs=mgf_s=mg, and this requires μF≥mg\mu F\ge mg.

Step 3. The minimum force is therefore Fmin=mgμF_{min}=\dfrac{mg}{\mu}.

Step 4. Since the coefficient of friction μ\mu is, for essentially all real surface pairs, a fraction less than 1, Fmin=mg/μF_{min}=mg/\mu is necessarily greater than mgmg.

✓Final answer

(c) Greater than mg.

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