A couple of problems on motion and force

AI Thread Summary
The discussion centers on calculating the maximum deceleration rate for a truck carrying a 3200-kg crate, given a static friction coefficient of 0.65. The normal force is determined to be 31,360 N, leading to a maximum frictional force of 20,384 N. Participants emphasize using Newton's second law (F = ma) to relate the decelerating force to the crate's motion. The frictional force, which is crucial for preventing the crate from crushing the cab, is highlighted as the key factor in determining safe deceleration. The conversation concludes with a focus on deriving the maximum deceleration and minimum stopping distance based on the calculated frictional force.
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Homework Statement



A flatbed truck is carrying a 3200-kg crate of heavy machinery. If the coefficient of static friction between the create and the bed of the truck is 0.65, what is the maximum rate at which the driver can decelerate when coming to a stop in order to avoid crushing the cab with the crate?

Homework Equations



fric = u*N
Fnet = m * a

The Attempt at a Solution



N = 3200*9.8 = 31360
fric = 0.65 * 31360 = 20384 N

thats as far as I got
 
Last edited:
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thechuckisback said:
fricc= u*N*d*e I think

What are d and e? The frictional force is equal to the coefficient of friction, mu, times the normal force which is usually denoted by N.
 
For 1), use Newton's 2nd law, F = ma. F is the decelerating force and the same force F acts on both the truck and the crate. But the actual force that slows down the crate is the force of friction between the crate and the floor of the truck. So what is the maximum frictional force you can have ?

2) is similar. Use F = ma and Fr = μm to get a. And minimum stopping distance comes from maximum decelerating (frictional) force.
 

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