How Does Cart Acceleration Affect Tension in a Moving Pendulum?

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The discussion focuses on calculating the tension in a cable connecting a moving cart and a pendulum load in a gantry crane model. The user has developed a formula for tension but is uncertain about its completeness, specifically regarding the impact of the cart's acceleration on tension. They propose that increased acceleration leads to greater tension due to higher angular velocity, but question whether an additional term for the driving force is necessary. Suggestions from participants include refining the tension formula to account for both gravitational and centripetal forces. The conversation emphasizes the relationship between cart acceleration and cable tension in the pendulum system.
phlegmy
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hey guys
i'm writing a computer prog in visual basic that models a gantry crane.
(basically modeling a pendulum that has a moving point of suspension)

the point of suspension, the cart, has a mass
the bob of the pendulum, the load, has a mass also

in order to calculate the net forces (horizontal) on the cart and load, to determine their accelerations, velocities and displacements, it is nescessary to know the tension in the cable that connects them.

what i need is a formula for calculating the tension in the cable given the two masses , the length of cable and the driving force F on the cart!

so far i have gotten the programme to work but am worried that my formula for tension is not complete

i have taken the tension= loadmass*g*cos(angle) + loadmass*cablelength*angularvelocity
, where the angle is the angle incurred by the cable to the verticle
which i think is correct for a pendulum with a fixed point of suspension
the second term is the centripedal force due to angular velocity

the net horizontal force on the load = tension*sin(angle)
and net horizontal force on cart = F- tension*sin(angle)



has anyone got any suggestions
i can "picture" it in my head;
the faster i accelerate the cart (very large F)
the greater the tension in the cable
is this greater tension accounted for in the fact that if the cart accelerates very quicly, the anglular velocity will be much greater and as a result tension will be much greater?
OR is it nescessary (as i suspect) to add another term containing F, the driving force on the cart, to the formula for tension?

any suggestions or insights appreciated
james!
 
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tension on cable = mg*(cos(angle of cable with normal)) + m*(v^2)/(cable length)

Force on mass= mg*(sin(angle of cable with normal)
 

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