Refer to the diagram shown below.
The hoist is in static equilibrium supported by tensions in the two ropes.
For horizontal force balance, obtain
T₃ cos 50 = T₂ cos 38
0.6428T₃ = 0.788T₂
T₃ = 1.2259T₂ (1)
For vertical force balance, obtain
T₂ sin 38 + T₃ sin 50 = 350
0.6157T₂ + 0.766T₃ = 350 (2)
Substitute (1) into (2).
0.6157T₂ + 0.766(1.2259T₂) = 350
1.5547T₂ = 350
T₂ = 225.124 N
T₃ = 1.2259(225.124) = 275.979
Answer:
T₂ = 225.12 N
T₃ = 275.98 N
Answer:
Explanation:
Givens
Vi = 10 m/s
Vf = 40 m/s
a = 3 m/s^2
Formula
a = (vf - vi) /t Substitute the givens into this formuls
Solution
3 = (40 - 10) / t Multiply both sides by t
3*t = t(40 - 10)/t Combine. Cancel t's on the right
3*t = 30 Divide by 3
3t/3 = 30 / 3
Answer: t = 10 seconds.
Answer: a. 198.6J b. - 198.6J
Explanation: Parameters given:
m = 15kg
g = 9.8m/s²
∅ = 12°
a. Work done by the force Fp on the cart if the ramp is 6.5m long.
Given the formula, Fp = Mgsin∅ = 15 x 9.8 x sin12° = 30.56N
Therefore Work done (Wp) = Fp x Ramp Length = 30.56 x 6.5 = 198.64Nm or 198.6J
b. The work done by the force mg on the cart.
Since the cart is being pushed upwards, it acts against gravity with its direction of motion. Taking into account the formula from the previous answer for Work Done (Wg) = Fmg x distance
= 15kg x -9.8m/s² x Sin12° x 6.5m
= - 198.6J
Vo = 5.89 m/s Y = 1.27 m g = 9.81 m/s^2
Time to height
Tr = Vo / g Tr = (5.89 m/s) / (9.81 m/s^2) Tr = 0.60 s
Max height achieved is:
H = Vo^2 / [2g] H = (5.89 )^2 / [ 2 * (9.81) ] H = (34.69) / [19.62] H = 1.77 m
It falls that distance, minus Andrew's catch distance:
h = H - Y h = (1.77 m) - (1.27 m) h = 0.5 m
Time to descend is therefore:
Tf = √ { [2h] / g ] Tf = √ { [ 2 * (0.5 m) ] / (9.81 m/s^2) } Tf = √ { [ 1.0 m ] / (9.81 m/s^2) } Tf = √ { 0.102 s^2 } Tf = 0.32 s
Total time is rise plus fall therefore:
Tt = Tr + Tf Tt = (0.60 s) + (0.32 s) Tt = 0.92 s (ANSWER)
