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II. Short Answer Questions · Q5

Q.Using a free body diagram, show that it is easier to pull an object than to push it.

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Step 1. When an object is PUSHED at angle θ\theta to the horizontal, the applied force FF has components Fsin⁡θF\sin\theta (parallel to the surface) and Fcos⁡θF\cos\theta (into the surface, downward). Vertical equilibrium gives Npush=mg+Fcos⁡θN_{push}=mg+F\cos\theta.

Step 2. The maximum static friction available is then fs,max=μsNpush=μs(mg+Fcos⁡θ)f_{s,max}=\mu_sN_{push}=\mu_s(mg+F\cos\theta) — larger than μsmg\mu_smg, so pushing increases the friction that must be overcome.

Step 3. When the SAME object is instead PULLED at the same angle θ\theta (force directed up and away from the surface), the vertical component Fcos⁡θF\cos\theta acts upward, partly supporting the weight: Npull=mg−Fcos⁡θN_{pull}=mg-F\cos\theta.

Step 4. Hence fs,max=μsNpull=μs(mg−Fcos⁡θ)f_{s,max}=\mu_sN_{pull}=\mu_s(mg-F\cos\theta), smaller than in the push case. …

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