Concept Revolves around Friction H.You are trying to move a heavy packing case. You pull it and push it(Figures 52a and 52b respectively), and find that a smaller force is requiredto make it slide if you are pulling.(a) Explain this in qualitative terms only.(b) If the coefficient of friction between case and floor is u and the weightof the case is W, find an algebraic expression for the force H requiredto move the case when pulling.(a) Pulling(c) Find the equivalent expression when pushing. (Hint: the push is anegative pull.)(d) For W 1 kN, µ=0.5, and 0= 30° evaluate H for pushing and alsofor pulling.(b) PushingFiaure 52

Pulling and pushing of a heavy packing case, the force required to slide it is smaller while pulling. (a) Explanation; **According to guidelines only three sub-parts can be answered, resubit the question to get answer of remaining sub-part(s).** The FBD of the of forces will clear that why it requires less force while pulling, While pulling; As friction force is Ff=μN And here the N<W , because the vertical component of the pulling force (H) is helpful in reducing the Weight (W) effect. While pushing; As friction force is Ff=μN And here the N>W , because the vertical component of the pulling force (H) is helpful in increasing the Weight (W) effect. **Hence as friction reduces while pulling that is why the pulling force is smaller** (b) Expressing of H while pulling; ∑Fx=0Ff=Hcosθ ...i∑Fy=0N+Hsinθ=W ...ii From eq (i) and (ii) we get; μW-Hsinθ=HcosθμW=Hcosθ+μHsinθμW(cosθ+μsinθ)=H(c) Expressing of H while pushing; ∑Fx=0Ff=Hcosθ ...i∑Fy=0N=Hsinθ+W ...ii From eq (i) and (ii) we get; μW+Hsinθ=HcosθμW=Hcosθ-μHsinθμW(cosθ-μsinθ)=H