Statika Zadaci Za Srednju Skolu Fixed Jun 2026

This is one of the most fundamental types of problems. Concurrent forces are forces whose lines of action intersect at a single point. The equilibrium condition is that the vector sum is zero: ΣF = 0. A typical task might involve:

FA−40=0⇒FA=40 kNcap F sub cap A minus 40 equals 0 implies cap F sub cap A equals 40 kN

Ako sila okreće telo u smeru kazaljke na satu, momentu se često dodeljuje znak (ili obrnuto, zavisno od usvojene konvencije).

30⋅4+30⋅1.5−27.5⋅6=120+45−165=165−165=030 center dot 4 plus 30 center dot 1.5 minus 27.5 center dot 6 equals 120 plus 45 minus 165 equals 165 minus 165 equals 0

G=m⋅g=10 kg⋅10 m/s2=100 Ncap G equals m center dot g equals 10 kg center dot 10 m/s squared equals 100 N

Korak 3: Postavljanje jednačine vertikalnih sila za nalaženje FAcap F sub cap A ∑Y=0sum of cap Y equals 0

M = 20 kN * 2,5 m = 50 kNm

(FB⋅4 m)−(Q⋅2 m)=0open paren cap F sub cap B center dot 4 m close paren minus open paren cap Q center dot 2 m close paren equals 0

Let’s walk through the problem that finally clicked for her.

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This is one of the most fundamental types of problems. Concurrent forces are forces whose lines of action intersect at a single point. The equilibrium condition is that the vector sum is zero: ΣF = 0. A typical task might involve:

FA−40=0⇒FA=40 kNcap F sub cap A minus 40 equals 0 implies cap F sub cap A equals 40 kN

Ako sila okreće telo u smeru kazaljke na satu, momentu se često dodeljuje znak (ili obrnuto, zavisno od usvojene konvencije). statika zadaci za srednju skolu fixed

30⋅4+30⋅1.5−27.5⋅6=120+45−165=165−165=030 center dot 4 plus 30 center dot 1.5 minus 27.5 center dot 6 equals 120 plus 45 minus 165 equals 165 minus 165 equals 0

G=m⋅g=10 kg⋅10 m/s2=100 Ncap G equals m center dot g equals 10 kg center dot 10 m/s squared equals 100 N This is one of the most fundamental types of problems

Korak 3: Postavljanje jednačine vertikalnih sila za nalaženje FAcap F sub cap A ∑Y=0sum of cap Y equals 0

M = 20 kN * 2,5 m = 50 kNm

(FB⋅4 m)−(Q⋅2 m)=0open paren cap F sub cap B center dot 4 m close paren minus open paren cap Q center dot 2 m close paren equals 0

Let’s walk through the problem that finally clicked for her. A typical task might involve: FA−40=0⇒FA=40 kNcap F