# Resolution of Forces and Center of mass

#### mm391

1. Homework Statement

A trailer with a mass of 900 kg with mass centre (centroid) at G is attached to a
towing car at A. If the car and trailer reach a velocity of 60 km/h along a horizontal road in a distance of 30 m from rest with constant acceleration, compute the vertical component of
the force at A. Assume the wheels are relatively light and that the friction force between
wheels and road can be neglected.

(Picture attached)

2. Homework Equations

SUVAT Equations

v^2=U^2+2as

3. The Attempt at a Solution

I am not sure where to even begin with this question. I worked out the acceleration to be 4.6m/s but have no idea were to go or how to finish. If someone could help it would be much appreciated.

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#### mm391

I think I worked it out.

Is it linear acceleration = 5.6m/s^2

Convert that to angular acceleration = 5.6/1.2 = 4.6rad/s

Than using trig work out the angle between A and G = 18.4°

then find the Y component of the acceleration and multiply it by 900kgs. which give you

Tan(18.4)*4.66 = 1.53

1.55 * 900 = 1395 Newtons?

#### PhanthomJay

Homework Helper
Gold Member
That answer appears correct, however, I have no idea why. There is no acceleration in the y direction, since the trailer does not lift off the ground. The linear acceleration of the center of mass is 4.6 m/s^2. There is no angular acceleration of the center of mass, or else it would tip over. Try solving for the net force in the horizontal direction, then sum moments about the center of mass of the horizontal and unknown vertical force at A.

#### mm391

I am still unclear about where to start. How do you sum the moments up in the x direction as the only think in the x-direction is 16.6m/s?

Then the moments about A = mg*1.2.........?
Moments about G = Reactions at A * 1.2....?

#### PhanthomJay

Homework Helper
Gold Member
I am still unclear about where to start. How do you sum the moments up in the x direction as the only think in the x-direction is 16.6m/s?

Then the moments about A = mg*1.2.........?
Moments about G = Reactions at A * 1.2....?
The final velocity in the x direction is given as 60 km/hr, or 16.67 m/s, after traveling 30m from rest. Using your relevant SUVAT equation you noted in post #1, you can now calculate the acceleration in the x direction, and from that, using Newton's 2nd law, you can calculate the net force in the x direction, which is applied at A.

Now you want to sum moments of all forces about the center of mass, at G, which must sum to zero for rotational equilibrium. The horizontal force at A produces a counterclockwise moment (horiz force at A times perpendicular distance from line of action of that force to G). So it must be balanced by a clockwise moment (vertical force at A times perp distnce from line of action of vert force to G). Other forces produce no moments. Solve for the vert force component at A.