A960 kg car is pulling a 310 kg trailer. together the car and trailer move forward with an acceleration of 2.00 m/s2. ignore any frictional force of air drag on the car and all frictional forces on the trailer.
(a) determine the net force on the car.
magnitude

n
direction

(b) determine the net force on the trailer.
magnitude

n
direction

(c) determine the force exerted by the trailer on the car.
magnitude

n
direction

(d) determine the resultant force exerted by the car on the road. (assume that the forward direction is along the +x-direction.)
magnitude

a)

b)

c)

d)          

    from the horizontal.

Explanation:

Given:

mass of the car,

mass of trailer,  

acceleration of the system,  

(a)

Net force on the car will be in the direction of acceleration given by:

(b)

(c)

Since the car and the trolley are linked together, the link will face a tension force and this will be exerted on the car by the Newton's third law of motion which can be given as:

(d)

Net vertical force acting on the road due to car:

Net vertical force acting on the road:

Now, the net resultant force:

Angle f the force from horizontal:

from the horizontal.

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.