# Calculating resultant forces given FBD

cdn88

## Homework Statement

A truck of mass 8000 kg is accelerating at 5.0 m/s2 up a hill of inclination 4 degrees to the horizontal The friction forces on the truck add up to 5500 N.

What force does the truck exert against the road in the direction of travel?

What force does the road exert on the truck in the direction of travel?

With the truck's engine working just as hard, what would be the acceleration of the truck down the hill?

F=ma

## The Attempt at a Solution

For the force the truck is exerting i used F=ma so F=(8000)(5)=40 000N
I just don't believe it's that simple, do i need to break down the incline into x and y components? In which case I am not given enough information..

Homework Helper
Gold Member
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## Homework Statement

A truck of mass 8000 kg is accelerating at 5.0 m/s2 up a hill of inclination 4 degrees to the horizontal The friction forces on the truck add up to 5500 N.

What force does the truck exert against the road in the direction of travel?

What force does the road exert on the truck in the direction of travel?

With the truck's engine working just as hard, what would be the acceleration of the truck down the hill?

F=ma

## The Attempt at a Solution

For the force the truck is exerting i used F=ma so F=(8000)(5)=40 000N
I just don't believe it's that simple, do i need to break down the incline into x and y components? In which case I am not given enough information..
you are mis-stating Newton's 2nd law. It's F_net =ma. F_net consists of several forces. What are they? You have all the info you need.

cdn88
tyou are mis-stating Newton's 2nd law. It's F_net =ma. F_net consists of several forces. What are they? You have all the info you need.

The forces are normal, gravity, friction, and the trucks excretion on the road, normal and gravity cancel, so i need to incorporate friction?

another thought, 5500N cos 4 and 5500N sin 4 give me x and y components of friction, I am just thinking about how to apply this to the equation.

Last edited:
Homework Helper
Gold Member
The forces are normal, gravity, friction, and the trucks excretion on the road,
Well, Yes!,but when you are looking at the forces on the truck,you are looking at the force exerted by the road on the truck, the Earth on the truck,the normalforce on the truck,and the friction on the truck
normal and gravity cancel,
No!
so i need to incorporate friction?
Yes, and the gravity component down the plane.
another thought, 5500N cos 4 and 5500N sin 4 give me x and y components of friction, I am just thinking about how to apply this to the equation.
When you are dealing with problems involving inclines, it is most often best to choose the x-axis parallel to the incline, and the y-axis perpendicular to the incline. It makes life a lot easer. You just have to tilt your head and watch your geometry and trig. So you have the truck force along the x direction parallel to and up the incline, and the friction force along the x-axis parallel to and down the incline; now you just need to know the component of the gravity force down the incline,and use Newton 2 to solve for the force exerted by the road on the truck. Then what about the force exerted by the truck on the road??
And welcome to PF!

cdn88
ok, thank you! so i set up the incline to be the xaxis. now, the force of gravity is found by f = mg ((8000)(9.8) = 78400 so 78400 sin 4 = 5468.9 which is the x component of the gravity force?

so, does f_net=ma mean 40000 is the total force so i subtract gravity force and friction to find what the truck is exerting to move 5m/s up the incline?

if so, 40000-5500-5441= 29059 N am i on the right track?