Physics, 04.08.2019 16:00 Evixie84

In order to find the magnitude of force f, we have to solve a system of two equations with both f and the normal force n unknown. use the methods we have learned to find an expression for f in terms of m, g, θ, and μs (no n).

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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, $\mu_s$ is presented as follows;

$Force, F = \mathbf{ \dfrac{\mu_s \cdot m \cdot g }{{cos\theta +sin\theta \cdot \mu_s } }}}$

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, F_g = m × g

The normal force = N (upwards direction)

The static friction force, f_s = $\mu_s$ × N

The two separate equations in F and N are;

F·cos(θ) - $\mu_s$ ·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 $\mu_s$

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)/ $\mu_s$

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)/ $\mu_s$ = m·g - F·sin(θ)

F·cos(θ)/ $\mu_s$ + F·sin(θ) = m·g

F·(cos(θ)/ $\mu_s$ + sin(θ)) = m·g

$F = \dfrac{m \cdot g }{\dfrac{cos\theta}{\mu_s} + sin\theta } = \dfrac{m \cdot g }{\dfrac{cos\theta +sin\theta \cdot \mu_s }{\mu_s} }} = \mathbf{\dfrac{\mu_s \cdot m \cdot g }{{cos\theta +sin\theta \cdot \mu_s } }}}$

Therefore;

$Force, F = \mathbf{ \dfrac{\mu_s \cdot m \cdot g }{{cos\theta +sin\theta \cdot \mu_s } }}}$

Learn more equilibrium of forces here:

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In order to find the magnitude of force f, we have to solve a system of two equations with both f an

Answer from: Lilyy1k

An expression for f in terms of m, g, θ, and μs $\large{\boxed{\bold{F~=~\frac{{mg-\frac{fs_{max}}{\mu~s} } }{{sin\:\theta}}}}$ Further explanation

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

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Answer from: Quest

Answavelength; amplitude;

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