Prof Sulieman Tarawn Stotics First Exam 0 Taculty 1 d Q isR Tha resultant of two

Question

Prof Sulieman Tarawn Stotics First Exam 0 Taculty 1 d Q isR Tha resultant of two torces Pan IP Q is daubled the neuo resullant is L to P. End the inter relationship between P, G and R . rod AB Carries three loc and 100 N at distances of 2o mm ,9omm and 150 m respectiveby PromA. Find the balancing point of he Ra? 2-A ds 30N,70N, 3-4Move tharce into C, B and A lo b) Find the resultant R elative ) what is the c withRA 00 N B C 20 The motor at B winds up the cord attached to the 30-kg crate with a constant speed. The force in cord CD supports the pulley C and the angle represents the equilibrium state- )raw the free-body diagram of the pulley C. Neglect the size of the pulley. cB a 4) rind Reaction at D all te poiat way from Ba g suatair a ore of h distance of y - 1 m. The cords CAD and CBD are attached to the rinrawthe free-body diagrams for point A and ring C. 9 Find AX 40 NM "

Explanation / Answer

1).

Resultant of two forces P and Q is R

|R| = sqrt(P^2 + Q^2 + 2PQCos)

When Q is doubled new resultant R’ is

R’ = P + 2Q

Since it is perpendicular to P, dot product of R’ and P is zero

(R’).(P) = 0

(P + 2Q).(P) = 0

P^2 + 2PQCos = 0

(Dot product of P and P is P^2 and that of P and 2Q is 2PQCos)

Cos = -P/(2Q)

R = sqrt(P^2 + Q^2 + 2PQCos)

R = sqrt(P^2 + Q^2 + 2PQ*(-P/(2Q))

R = sqrt(P^2 + Q^2 - P^2)

R = Q

2).

Moment about A =( (10X150)+(7X90)+(3X20)) = 2190.
Total Load = 100+70+30 = 200
Eccentricity = 2190/200 = 109.5 mm

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