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DedPeter [7]
2 years ago
15

(15 Points)

Physics
1 answer:
oksian1 [2.3K]2 years ago
5 0

The vertical weight carried by the builder at the rear end is F = 308.1 N

<h3>Calculations and Parameters</h3>

Given that:

The weight is carried up along the plane in rotational equilibrium condition

The torque equilibrium condition can be used to solve

We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person

This would lead to:

F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)

F(1cos20)= 197/2(3.10sin20 + 2 cos 20)

Fcos20= 289.55

F= 308.1N

Read more about vertical weight here:

brainly.com/question/15244771

#SPJ1

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A force of 48 newtons is required to start a 5.0 kg box moving across a horizontal concrete floor. What is the coefficient of st
Soloha48 [4]

1) The coefficient of static friction is 0.980

2) The coefficient of kinetic friction is 0.908

Explanation:

1)

In the first situation, the box is still at rest. There are two forces acting on the box:

- The force of push, F, forward

- The force of static friction, F_f

Since the box is in equilibrium we have

F_f = F (1)

The value of frictional force changes from zero to a maximum value which is given by:

F_f = \mu_s mg (2)

where

\mu_s is the coefficient of static friction

m is the mass of the box

g is the acceleration of gravity

So, if the force F needed to put the box in motion is 48 N, it means that this is also the maximum value of the force of friction. So, we can combine eq.(1) and (2) to find the coefficient of static friction:

\mu_s = \frac{F}{mg}

where:

F = 48 N

m = 5.0 kg

g=9.8 m/s^2

Substituting,

\mu_s = \frac{48}{(5.0)(9.8)}=0.980

2)

In this second situation, the object is already in motion, so the equation of motion is:

F-F_f = ma

where

F = 48 N is the force applied forward

a = 0.70 m/s^2 is the acceleration of the box

F_f = \mu_k mg is the force of kinetic friction, where

\mu_k is the coefficient of kinetic friction

We can therefore rearrange the equation to find the coefficient:

F-\mu_k mg = ma\\\mu_k mg = F-ma\\\mu_k = \frac{F-ma}{mg}=\frac{48-(5.0)(0.70)}{(5.0)(9.8)}=0.908

Learn more about friction:

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