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erica715
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This is actually for an engineering dynamics class but its basically physics. I know how to find force using F=ma. But I can't figure out how to find tension given velocity and acceleration. My book is no help :(
1. A 3.5 Mg engine is suspended from a spreader beam AB having a negligible mas and is hoisted by a crane which gives it an acceleration of 4 m/s^2 when it has a velocity of 2m/s. Determine the force in chains CA and CB during the lift. To help you visualize AB represents the distance between where the two ropes connect to the top of the engine and C is where the two ropes meet together at the crane. The angle between line AC and AB is shown as 60 degrees.
F=ma
First I tried the obvious F=ma. 3500kg(9.81)(4)=137340. Wrong.
Then I tried...T-cos(60)(3500)(9.81)-sin(60)(3500)(9.81)=(3500)(4)
T=26567.48. Wrong.
I feel like I need to include the velocity that's given but I have no idea where.
The answer in the back of the book says that T(of CA) and T(of CB) are both 27.9 kN.
Ugh I've been at this for a couple hours now and I only did 2 of my homework problems so far. Help please!
1. A 3.5 Mg engine is suspended from a spreader beam AB having a negligible mas and is hoisted by a crane which gives it an acceleration of 4 m/s^2 when it has a velocity of 2m/s. Determine the force in chains CA and CB during the lift. To help you visualize AB represents the distance between where the two ropes connect to the top of the engine and C is where the two ropes meet together at the crane. The angle between line AC and AB is shown as 60 degrees.
Homework Equations
F=ma
The Attempt at a Solution
First I tried the obvious F=ma. 3500kg(9.81)(4)=137340. Wrong.
Then I tried...T-cos(60)(3500)(9.81)-sin(60)(3500)(9.81)=(3500)(4)
T=26567.48. Wrong.
I feel like I need to include the velocity that's given but I have no idea where.
The answer in the back of the book says that T(of CA) and T(of CB) are both 27.9 kN.
Ugh I've been at this for a couple hours now and I only did 2 of my homework problems so far. Help please!