A crate of mass m = 31 kg rides on the bed of a truck attached by a cord to the back of the cab as in the figure. The cord can withstand a maximum tension of 82 N before breaking. Neglecting friction between the crate and truck bed, find the maximum acceleration the truck can have before the cord breaks.
A dockworker loading crates on a ship finds that a 22-kg crate, initially at rest on a horizontal surface, requires a 80-N horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of 59 N is required to keep it moving with a constant speed. Find the coefficients of static and kinetic friction between crate and floor.
f= m . g(acceleration(a))
a=f/m
a= 82N/31 kg
=2.65 ms2
To find the maximum acceleration the truck can have before the cord breaks, we need to consider the tension in the cord.
Let's assume the acceleration of the truck is a, and the tension in the cord is T.
Using Newton's second law, we can write the force equation for the crate in the vertical direction:
T - mg = 0 [since there is no vertical acceleration]
Now, considering the horizontal direction, the net force acting on the crate is given by:
ma = T
We want to find the maximum acceleration (a) that the truck can have before the cord breaks, which occurs when the tension (T) in the cord reaches its maximum value of 82 N.
So we rewrite the force equation as:
ma = 82 N
Substituting the weight (mg) with its value:
ma = 82 N + mg
Since m = 31 kg and g ≈ 9.8 m/s²:
31a = 82 N + 31 kg × 9.8 m/s²
31a = 82 N + 303.8 kg·m/s²
31a = 82 N + 303.8 N
31a = 385.8 N
Divide both sides by 31:
a = 385.8 N / 31
a ≈ 12.476 m/s²
Therefore, the maximum acceleration the truck can have before the cord breaks is approximately 12.476 m/s².
To find the maximum acceleration the truck can have before the cord breaks, we need to consider the forces acting on the crate.
First, let's analyze the forces acting on the crate:
1. Weight (mg): The weight of the crate is given by the product of its mass (m) and acceleration due to gravity (g). Therefore, the weight is W = mg.
2. Tension (T): The tension in the cord is the force with which the truck pulls the crate. We need to find the maximum tension before the cord breaks. Let's denote the maximum tension as T_max = 82 N.
Next, let's consider the free-body diagram for the crate:
↑ T
|
|
<----|---->
m | W
|
Since the crate is not moving vertically, the vertical components of the forces balance each other:
T - W = 0
Therefore, we can write T = W.
Now, let's substitute the values into the equation:
T_max = W
T_max = mg
To find the maximum acceleration (a_max) that the truck can have before the cord breaks, we need to set up Newton's second law for the crate:
F_net = ma
In this case, the net force is the tension in the cord (T) and the mass is given. Therefore:
T - W = ma
Since T = W, we can substitute it into the equation:
W - W = ma
0 = ma
Since the weight (mg) is equal to the maximum tension (T_max) before the cord breaks, the acceleration is zero (a = 0). This means that the maximum acceleration the truck can have before the cord breaks is zero.
In practical terms, this means that the crate would not move as long as the tension in the cord is equal to or below the maximum tension (82 N). Any acceleration greater than zero would exceed the maximum tension and cause the cord to break.