ocr mechanics 1

Hello, I need some help with a maths problem please :smile:

A car of mass 1000kg is towing a camping trailor of mass 250kg along a straight road. there are constant resistances to the motion of the car and the trailer of magnitude 150N and 50N respectively. the driving force on the car has magnitude 800N. calculate the acceleration of the car and the trailer and the tension in the towbar when the road is horizontal.

I've worked out the acceleration , 0.48 m s^-2

But i have no idea how to work out the tension in the towbar.

Thanks in advance

Reply 1

Dream Eater

The resultant force on the car is 800-150-T, since it has a driving force of 800N, resistance 150N, and tension in the towbar T. The resultant force on the trailer is T-50, since only the tension of the bar is pulling it, and it has resistance 50N. Using newton's 2nd law F=ma:

800-150-T=1000a,
T-50=250a,

and you just solve simultaneously to get a=0.48, T=170 :smile:

.

Reply 2

mousy

OP

14

Dream Eater

The resultant force on the car is 800-150-T, since it has a driving force of 800N, resistance 150N, and tension in the towbar T. The resultant force on the trailer is T-50, since only the tension of the bar is pulling it, and it has resistance 50N. Using newton's 2nd law F=ma:

800-150-T=1000a,
T-50=250a,

and you just solve simultaneously to get a=0.48, T=170 :smile:

.

Thank you very much

Reply 3

steve2005

17

mousy

Tension in toolbar does two things. It accelerates trailer and overcomes resistance.
T= ma + R
T=250(0.48)+50=170N

Reply 4

mousy

OP

14

Okay, the next part of the question is exactly the same but the road is inclined at sin^-1 (2/49) to the horizontal with the car travelling uphill. How do i tackle this??

Reply 5

Dream Eater

mousy

Okay, the next part of the question is exactly the same but the road is inclined at sin^-1 (2/49) to the horizontal with the car travelling uphill. How do i tackle this??

Ok, so the question is exactly the same, except there is a component of gravity acting on both the car and the truck. Gravity pulls vertically downwards on the car with a force of 1000g, and on the truck with a force 250g (I'm guessing they've asked you to take g = 9.8, as that would explain the weird choice of angle). The gravity component of the force on the car is 1000g * sin(arcsin(2/49)). The gravity acting on the truck gives 250g * sin(arcsin(2/49)). Your new resultant forces are:

800-150-T-1000g*(2/49) = 1000a,
T-50-250g*(2/49) = 250a,

and you solve simultaneously again to get a=0.08, T=170 (maybe you can double-check that for me). Hope that helps! :biggrin:

Reply 6

mousy

OP

14

Yep, they are the correct answers :smile:

Thank you very very much :biggrin:

Reply 7

mousy

OP

14

I have another question i am stuck on, this time its about pulleys.

Two children, P and Q, of masses 40kg and 50kg respectively, are holiding on to the ends of a rope which passes over a thick horizontal branch of a tree. The parts of the rope on either side of the branch are vertical and child Q is moving downwards. A model is to be used in which the children may be considered as particles, and in which the rope is light and inextensible and is moving freely in a smooth groove on the branch. Show that the acceleration of each child has magnitude 1.09 m s^-2, and find the tension in the rope.

I can do this bit, I worked out the tension to be 436 newtons.
It is the next bit i am stuck on

When Child Q is moving at 2 metres per second, she lets go of the rope. Child P continues to rise for a further distance h metres before falling back to the ground. Calculate the value of h.

Thanks for any input :smile:

Reply 8

zrancis

2

Consider the child falling back down when her upwards velocity = 0 so she is at instantaneous rest. Using s=h, u=2, a=-g, v=0 should bring that out nicely for you

Reply 9

mousy

OP

14

zrancis

Consider the child falling back down when her upwards velocity = 0 so she is at instantaneous rest. Using s=h, u=2, a=-g, v=0 should bring that out nicely for you

thanks

Reply 10

Elle Jeffrey

Hello! I have a problem too if anyone could help me please!

Initiall a block of wood is at a point O on a rough plane inclined at 15degrees to the horizontal. The block is projected directly up the plane with initial speed 4ms-1. The coefficient of friction between the block and the plane is 1/10. the block instantaneously to rest at A. Find the distance OA.

ocr mechanics 1

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