What is the Maximum Safe Upward Acceleration of a Block Lifted by a Crane?

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Homework Help Overview

The discussion revolves around determining the maximum safe upward acceleration of a concrete block being lifted by a crane, given the crane's maximum safe working load and the block's mass. The subject area includes dynamics and forces, particularly focusing on the relationship between force, mass, and acceleration.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants attempt to apply Newton's second law (F=ma) to find the acceleration, but there is a recognition that the net force must account for gravitational force acting on the block. Some participants question the initial calculations and suggest reconsidering the effects of gravity on the net force.

Discussion Status

The discussion includes attempts to clarify the correct approach to calculating the net force and acceleration. Some participants have provided calculations that lead to a revised understanding of the problem, indicating a productive exploration of the topic, though no consensus has been reached on the final answer.

Contextual Notes

Participants note the importance of considering gravitational force in the calculations, which affects the net force available for acceleration. The original poster's calculations are challenged, highlighting the need for careful consideration of all forces involved.

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Homework Statement


A crane has a maximum safe working load of 1.2 * 10^4 N and is used to lift a concrete block of mass 1000kg. What is the maximum safe upward acceleration of the block while being lifted?

Homework Equations


F=ma

The Attempt at a Solution


F=mα
1.2*10^4=1000*α
α= 12m/s2The answer is 2.2 m/s2
 
Last edited:
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raamishstuden said:

Homework Statement


A crane has a maximum safe working load of 1.2 * 10^4 N and is used to lift a concrete block of mass 1000kg. What is the maximum safe upward acceleration of the block while being lifted?


Homework Equations


F=ma


The Attempt at a Solution


F=mα
1.2*10^4=1000*α
α= 12m/s2


The answer is 2.2 m/s2

You seem to have forgotten that while the crane is pulling up, gravity is pulling down, so the net force is much less than 1.2*10^4 N
Acceleration is the result of the net force.
 
Oh! I got it. Here is the solution. Is it right?

F-f=ma
(1.2*10^4)-(9.81*1000)=1000*a
2190=1000a
a=2.19
=2.2 ms-2
 
raamishstuden said:
Oh! I got it. Here is the solution. Is it right?

F-f=ma
(1.2*10^4)-(9.81*1000)=1000*a
2190=1000a
a=2.19
=2.2 ms-2

That's more like it.
 

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