taller de fisica

Nombres: Curso: Ing 29
Chang Zambrano AnaProfesor: Richard Piloso
Ferruzola Sánchez Lissette

Solución:






∑Fx=0 ∑Fx = Ta Cos 300 TA Cos 600 =0 ∑Fy= Ta Sen 300 + TA Sen 600T=0
∑Fy=0 Ta Cos 300 = TA Cos 600 Ta Sen 300 +TA Sen 600 = T
Ta = TA Cos 600 /Cos 300 T= 1154,7 N
Ta =577,4N
∑Fy= T – 50g-F =0
T – 50g =F
F= 1154,7 – 50(9.8)
F= 664,7 N

Solución:






Del diagrama BDel diagrama A ∑Fy = N- 30g = 0
∑Fy = 0 ∑Fx = 0 N = 30g
∑Fy = T – 20g =0∑Fy = 0 ƒSmáxima = 20g = µsN
T= 20g ∑Fx = T – ƒSmáxima = µsN20g = µs 30g
µs = 2/3 = 0,67

Solución:



∑Fx = 0 ∑Fy = 0 T =µs mAg·CosӨ·mAg ·SenӨ
ƒSmaxima · mAg ·SenӨ ·T = 0 N · mAg CosӨ = 0 T = mAg(µsCosӨ·SenӨ)
ƒSmaxima =µs mAg·CosӨ N = mAg CosӨT = 50(9.8) [(0.45) Cos200·Sen200) ]
T = 39.6 N



Solución:






∑Fx = 0 F = mg tanӨ
F CosӨ·mgSenӨ = 0 F = 20(9.8) tan400
F CosӨ = mgSenӨ...