What Is the Horizontal Force Between the Truck Tires and Ground?

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SUMMARY

The horizontal force between the truck tires and the ground is calculated to be 1392.5 lb based on the truck's weight of 5000 lb and an acceleration of 1 ft/s². The calculations involve separating the truck and the boulder into distinct free body diagrams (FBDs) and applying Newton's second law (∑F = ma). However, the expected answer of 765 lb suggests an oversight regarding friction, which is essential for the truck's movement. The discrepancy indicates a potential error in the provided solution rather than in the calculations performed.

PREREQUISITES
  • Understanding of Newton's second law (∑F = ma)
  • Ability to create and analyze free body diagrams (FBDs)
  • Basic knowledge of weight and mass conversion (using 32.2 ft/s² for gravitational acceleration)
  • Familiarity with the concept of tension in pulley systems
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  • Learn about tension forces in pulley systems
  • Study advanced applications of Newton's laws in multi-body systems
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rdg29
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1. Problem statement:

A 5000-lb truck is being used to lift a 1000-lb boulder B that is on a 200-lb pallet. Knowing the acceleration of the truck is 1 ft/s^2, determine the horizontal force between the tires and ground. The truck is on a horizontal plane with the boulder and pallet connected to a single pulley system off a cliff face.

Homework Equations

:[/B]

∑F = ma

bsCTqRi.jpg


3. Attempt at the solution:

[STRIKE]Assuming no friction.
[/STRIKE]
First, I separated the truck and the boulder with two separate FBD/Kinematic diagrams.

For the truck, I had 4 total forces acting on the body.

F = ? (horizontal force between the truck and the ground)
T = ? (tension in the cable)
N = 5000 lb
W = -5000 lb

Setting up the sum of the forces on the truck, we get:
∑Fy = 0
∑Fx = F - T
(ma)sum = F - T

mtruck = 5000/32.2 = 155.280

Subbing in the known values:
155.280*1 ft/s^2 = F - T

Now, I moved to the pulley system:

The FBD consisted of 3 forces:

T = ? (tension in the cable)
Wb = Weight of the Boulder = 1000 lb
Wp = Weight of the Pallet = 200 lb

Setting up the sum of the forces, we get:

∑Fx = 0
∑Fy = T - Wb - Wp

Subbing in the known values:
mboulder+pallet = (1000 lb + 200 lb)/32.2 = 37.267

(mboulder+pallet*1 ft/s^2) = T - 1000-lb - 200-lb
T = 1237.267-lb

Use T to find F in the equation for the truck:
155.280 = F - 1237.267
F = 1392.5-lb

The problem I am running into however, is this does not match the answer I'm told to find. I'm wondering if it is just something simple that I'm missing here, so any guidance would be appreciated.
 
Last edited:
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Looks ok to me except you should found off the number. There is friction,though, between driving tires and road. Otherwise, the truck couldn't move.
 
Thanks for looking it over, and pointing out the friction error on my part. The overall error may end up being an error in the solution provided maybe? The solution listed is 765-lb for the force between the ground and the truck.
 

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