Applying Newtons Second Law to Vertical Motion

This is not the answer to the original question, which asked for what force the cable exerts to give the car its acceleration.
  • #1
Ahmad786
17
0
18. An elevator that weighs 3.5 x 103 N is accelerated upward at 1.0 m/s2. What force does the cable exert to give it this acceleration?
A. 357 N
B. 0 N
C. 3500 N
D. 3857 N
(For this quesion I was really confused and I guessed that the force of tension in the cable was what the cable weighed )
Is that answer correct if not how do I find the correct answer?
 
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  • #2
Ahmad786 said:
(For this quesion I was really confused and I guessed that the force of tension in the cable was what the cable weighed )


This would only be true if the forces on the elevator car were balanced. However, since the elevator car has a non-zero acceleration, Newton's Second Law tells you that the forces on the elevator car are NOT balanced i.e. there is a net force acting. Hint: you can use Newton's Second Law to calculate the net force from the mass and acceleration, and then use the result to figure out how the tension compares to the weight.

Draw a free body diagram: it will help you immensely.
 
  • #3
"What force does the cable exert to give it this acceleration?" is this asking for net force if it is could I use the equation Fnet=Ft+Fg and If I used this Would it look like this?:
Fnet=Ft+Fg Fnet=(3500N/9.81m/s2)(1.00m/s2)+3500N so would the answer be D. 3857 N
 
  • #4
The car has two forces acting on it. Tension upward, and weight downward. The fact that the car is accelerating upward tells you that the tension must be greater than the weight (in magnitude). The net force is of course just the difference between the two (in other words, it is the amount by which the tension is greater than the weight). Therefore, your answer is correct, but your labelling of quantities is wrong. You are calculating Ft, not Fnet, using the formula:

Fnet = Ft - Fg

which implies: Ft = Fnet + Fg

which is the calculation you did, leading to an answer of D (for Ft).
 
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