Question

A 1020-kg car is on an inclined plane inclined at an angle of 30 degrees. A cable is connected to the car, stretched over a pulley at the end of the incline and connected to a 11900-N counterweight. See diagram. Acceleration up the plane is 3.09 m/s/s


(b) Determine the tension (in Newtons) in the cable.



(c) What mass (in kg) should the counterweight have in order for the car to move down the incline at an acceleration of 2.43 m/s/s?

**I give medals*** please help

Answer 1

(B)
For case of counterweight: mg-T=ma, where g is acceleration of gravity, a is acceleration, T is tension. Hence 11900-T=1020*3.09, then T= 8748.2N.
(C)
For the case of car F-R=ma, where F is driving force or force due to car's weight (Fw) and R is Resistive force, which is that of Tension
mgsin$-T=ma, because there is no friction, therefore no Frictional force.
1020*9.81*sin30-T=1020*2.43 and T=2524.5N
For the counterweight's case:
mg_T=ma or 9.81m-2524.5=2.43m, then 7.38m=2524.5, and m=342.1Kg.

Answer 2

i did the same thing for B but it says the answer is wrong. I thought it was because it requires a hundreths place decimal but there is none, so idk what's wrong D:

Answer 3

2.43 you wrote in your question yourself on C.

Answer 4

hahah oh my goodness yeah ok

Answer 5

it still says the answers are wrong though

Answer 6

Ok. Is the answer for B 8748.2sin30=4374.1N?

Answer 7

no D:

Answer 8

the only difference I see is for the way u solved C we only use 9.8 in our class not 9.81 and i tried fixing it with that but i got crazy numbers

Answer 9

Is it 8154.9N?

Answer 10

YES
!!!

Answer 11

I have made a minor mistake. Go back to my working on b and instead of mg-T=ma it should have been T-mg=ma, because if it is accelerating up the plane, then t is the driving force and Fw is the resitive one.

Answer 12

aaahhhh ok

Answer 13

thank you

Answer 14

You are welcome.

Answer 15

:)

Answer 16

(: