Suppose the worker in Exercise 6.3 pushes downward at an angle of 30 below the h
ID: 2035056 • Letter: S
Question
Suppose the worker in Exercise 6.3 pushes downward at an angle of 30 below the horizontal. (a) What magnitude of force must the worker apply to move the crate at constant velocity? (b) How much work is done on the crate by this force when the crate is pushed a distance of 4.5 m? (c) How much work is done on the crate by friction during this displacement? (d) How much work is done on the crate by the normal force? By gravity? (e) What is the total work done on the crate?
The following question is from University Physics chapter 6 exercise 6.4. I have attempted this problem multiple times but get it wrong because I keep using the wrong angle. The problem specifically mentions that the applied force is at an angle below the horizontal. Which I assume to be -30 degrees. However, the correct solution comes from using positive 30 degrees. Here is the example of the resulting free-body diagram:
Why is the angle positive and not negative? Why can I not use -30 or 330 degrees?
Fcos 30° Fsin 30 mgExplanation / Answer
Ans)
Given data is ?=30o
From exercise 6.3 mass m=30kg
Distance d=4.5m
Coefficient friction ?=0.25
The newtons second law states that the force is equals to the rate of change in momentum .
It also states that the net force acting on the body must be zero when the system is in the equilibrium .
The kinetic friction is defined as the amount of retarding force between two bodies that are moving relative to each other .
The expression is given as
f=?kN
a)what magnitude of force must the worker apply to move the crate at constant velocity
since the velocity is constant the acceleration is zero
Fcos(A)-?N=0
Fsin(A)+mg-N=0
Substitute N= Fsin(A)+mg
Fcos(A)-?(Fsin(A)+mg)=0
F(cos(A)-?sin(A))=?mg
F= ?mg/( cos(A)-?sin(A))
F=(0.25x30x9.8)/(cos30-0.25sin30)
F=(73.5)/(0.866-0.125)
F=99.2N
b)work done
W=99.2xcos(30)x4.5
=99.2x0.866x4.5
W=386.6 J
c)how much work is done on the crate by friction during this displacement
that is since 386.6 J is the work done against friction
therefore Wf=386.6J
d)how much work done by the normal force
Wnf=Nx0
Wnf=0J
e)total workdone
W=386.6J
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