On an object, there is a resultant force
Four styles usually work on objects, including:
gravity force, normal force and friction force
Gravity force itself is expressed as weight
While the normal force is a force acting on the touch plane of two surfaces and the direction is perpendicular
Friction force is the force that occurs when two objects touch each other
The direction of friction is opposite to the direction of motion
In the event of friction, there is a static friction force (fs), which is the frictional force at the position of a stationary object and the kinetic friction force acting when the object is moving (fk).
The equation is stated in
fs = us.N
fk = uk.N
us = static friction
uk = kinetic friction
If we want to express the force of push/pull by including μs, then the condition of the object is right to going to move
When in a state of silence, the force works :
In the vertical direction
N + Fy-mg = 0 ...(equation 1)
Fy = force in the direction of the y-axis
Fy = F sin θ
While N is obtained from:
fsmax = μs.N (fsmax = when the object starts to move)
So the equation 1 becomes:
fsmax /μs + F sin θ -mg = 0
F sin θ = mg - fsmax / μs
F = (mg-fsmax /μs) / sin θ
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Keywords: Friction, resultant force, weight
The magnitude of force acting on a body can be determined by using a system of simultaneous equations based on the sum of forces acting on a body in equilibrium is zero
The expression for the force, F, in terms of m, g, θ, [tex]\mu_s[/tex] is presented as follows;
[tex]Force, F = \mathbf{ \dfrac{\mu_s \cdot m \cdot g }{{cos\theta +sin\theta \cdot \mu_s } }}}[/tex]
Reason for arriving at the above expression is as follows;
Question; Part of the question that appear missing are;
The mass of the block = 10 kg
The force pulling the mass = F
The angle of elevation of the direction of the force, θ = 30°
The force of gravity, [tex]F_g[/tex] = m × g
The normal force = N (upwards direction)
The static friction force, [tex]f_s[/tex] = [tex]\mu_s[/tex] × N
The two separate equations in F and N are;
F·cos(θ) - [tex]\mu_s[/tex]·N = 0
F·sin(θ) + N - m·g = 0
The equations above are based on the equilibrium of forces acting on the mass being pulled by a force directed at an angle θ above the horizontal
Required;
Finding an expression for F in terms of m, g, θ, and [tex]\mu_s[/tex]
Method;
Make N the subject of the given system of equations, and equate both expressions of N as follows;
Finding the value of N based on the first equation, gives;
N = F·cos(θ)/[tex]\mu_s[/tex]
Finding the value of N based on the second equation, gives;
N = m·g - F·sin(θ)
Equating both values of N gives;
F·cos(θ)/[tex]\mu_s[/tex] = m·g - F·sin(θ)
F·cos(θ)/[tex]\mu_s[/tex] + F·sin(θ) = m·g
F·(cos(θ)/[tex]\mu_s[/tex] + sin(θ)) = m·g
[tex]F = \dfrac{m \cdot g }{\dfrac{cos\theta}{\mu_s} + sin\theta } = \dfrac{m \cdot g }{\dfrac{cos\theta +sin\theta \cdot \mu_s }{\mu_s} }} = \mathbf{\dfrac{\mu_s \cdot m \cdot g }{{cos\theta +sin\theta \cdot \mu_s } }}}[/tex]
Therefore;
[tex]Force, F = \mathbf{ \dfrac{\mu_s \cdot m \cdot g }{{cos\theta +sin\theta \cdot \mu_s } }}}[/tex]
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