(1) A flowerpot (m 10 kg) hangs from three wires as shown. Compute the tension i
ID: 1393323 • Letter: #
Question
(1) A flowerpot (m 10 kg) hangs from three wires as shown. Compute the tension in each wire. (Hint: To find T3, draw a free-body diagram for the flowerpot itself. To find T1 and 12, draw a free-body diagram for the knot where the three wires are joined.) 35 55 m 10 kg (2) What is the acceleration a of the cart (of mass MD? Assume both pulleys are massless, and there is no friction. Note that, because the hanging mass is suspended from a movable pulley, m moves half as far as M, so the acceleration of the hanging mass is a/2.Explanation / Answer
(1) Given that,
m = 10 kg,
theta1 = 55 degrees and theta2 = 35 degrees
We need to find T1, T2 and T3
From free body diagram, T3 can be simply obtained as,
T3 = W = mg = 10 x 9.8 = 98 Newtons
Now, the system is in equillibrium, so the net forces sums to zero,
F(net) = T1 + T2 + T3 = 0
Net force along x will be
Fx(net) = T1x + T2x+ T3x = 0
Fx(net) = -T1 cos(theta1) + T2 cos(theta2) + T3 cos(90) = -T1 cos 55 + T2 cos 35 = 0
T1 = T2 cos 35 / cos 55 = 1.43 T2
T1 = 1.43 (T2) eq(1)
Along Y, the net force will be
Fy(net) = T1y + T2y +T3y = 0
Fy(net) = T1 sin55 + T2 sin 35 + T3 sin 90 = 0
T1 sin 55 + T2 sin 35 + 98 = 0
using eq (1)
1.43 x T2 x 0.82 + T2 x 0.57 + 98 = 0
1.745 T2 = -98 => T2 = 56.2 N
T1 = 1.43 x 56.2 = 80.4 N
Hence,
T1 = 80.4 N ; T2 = 56.2 N and T3 = 98 N.
(Multiple questions asked, solved one, pls put the others seperately... thanks)
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