Dynamics Question: Solving for Force in a Frictionless Ramp Scenario

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Homework Help Overview

The discussion revolves around a dynamics problem involving a flat-bed truck with a frictionless ramp and a 5kg block resting on it. The scenario includes determining the force exerted by the truck's engine while considering the block's stationary position relative to the ramp.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the necessity of starting with free body diagrams and applying Newton's second law. There is uncertainty about the initial acceleration value and its relevance to the problem. The original poster attempts to calculate various forces and accelerations but questions the validity of their results.

Discussion Status

Some participants have provided guidance on starting the problem with free body diagrams. The original poster has revised their acceleration calculation, indicating an evolving understanding of the problem. However, there is no explicit consensus on the final answer or method yet.

Contextual Notes

Participants are navigating through assumptions about the relationship between the truck's and block's accelerations, as well as the implications of the friction coefficient in their calculations.

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Dynamics Question

Homework Statement



1. Homework Statement
A flat-bed truck with a frictionless ramp accelerates forward.


(That 35 is the angle, and a 5kg block rests on the hypotenuse)


/ |
--- / l
- - /35deg l
- ---------------
-----------------
-o-----------0--

The acceleration of the truck is suck that a 5.0kg block is stationary (relative to the ramp). If the rolling coefficient of friction is (mew=0.10) Find the force the engine is exerting on the 12000kg truck.


Homework Equations


F=ma Ff=mewFn Fg=mg


The Attempt at a Solution


Well...frankly, I really don't know how to start this question. How do I know where to start first? I'm wondering if I need acceleration in this or not, but I got a=0.98 m/s2 assuming that the truck's accel and the box's are the same.

Ok, so I reattempted it finding Fparallel as 40.138N assuming Fparallel is = to Fn
I found Ff as 4.0138N, and looked for Fperpendicular which = to 28.105N
I subracted Ff from Fperpendicular to find Fnet which I got as 24.09.
I took the Fnet and divided by the mass of the crate which is 5kg on the ramp which got me 4.82m/s2

Assuming that the acceleration of the box, and the acceleration of the truck to be the same,
I did Fnet=ma which is Fnet=12000 x 4.82 = 56840N

It's a pretty big number, and I'm wondering if anyone can help me confirm if I did it right.
 
Last edited:
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Faylight said:

The Attempt at a Solution


Well...frankly, I really don't know how to start this question. How do I know where to start first?

You always start this sort of problem by drawing free body diagrams and applying Newton's second law to them. Look at the examples in your book and notes. You should see this.

I'm wondering if I need acceleration in this or not, but I got a=0.98 m/s2 assuming that the truck's accel and the box's are the same.

It's true that the truck and the box have the same acceleration, but that common acceleration is not 0.98 m/s2. How did you get this?
 
For the acceleration, I got that wrong so I redid it, and I ended up with a new acceleration of 5.26m/s2
 
hi, i was wondering if you figured out the final answer to this question?
 

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