A 250-kg crate rests on a surface that is inclined above the horizontal at an angle of 21.8°. A horizontal force (magnitude = 515 N and parallel to the ground, not the incline) is required to start the crate moving down the incline. What is the coefficient of static friction between the crate and the incline?
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Hello
The relevant point here is, that the component of the pulling force that is normal to the incline, reduces the normal force of the crate, and thus reduces the friction force.
The pulling force parallel to the incline is
Fp = 515*cos21.8 N
the pulling force normal to incline = 515*sin21.8 N
the gravitational force down the incline is
Fg = mgsin21.8 = 250*9.81*sin21.8 N
the normal force of the crate is
Fn = mgcos21.8 = 250*9.81*cos21.8
the friction force is
Ff = (normal force - perpendicular component of pulling force)* fric. coefficient
Ff = (250*9.81*cos21.8 - 515*sin21.8)*μ
Parallel component of pulling force + gravitational force = friction force
515*cos21.8 + 250*9.81*sin21.8 = (250*9.81*cos21.8 - 515*sin21.8)*μ
μ = (515*cos21.8 + 250*9.81*sin21.8) / (250*9.81*cos21.8 - 515*sin21.8)
μ = 1388.94/2085.85
μ = 0.6658
Regards
The relevant point here is, that the component of the pulling force that is normal to the incline, reduces the normal force of the crate, and thus reduces the friction force.
The pulling force parallel to the incline is
Fp = 515*cos21.8 N
the pulling force normal to incline = 515*sin21.8 N
the gravitational force down the incline is
Fg = mgsin21.8 = 250*9.81*sin21.8 N
the normal force of the crate is
Fn = mgcos21.8 = 250*9.81*cos21.8
the friction force is
Ff = (normal force - perpendicular component of pulling force)* fric. coefficient
Ff = (250*9.81*cos21.8 - 515*sin21.8)*μ
Parallel component of pulling force + gravitational force = friction force
515*cos21.8 + 250*9.81*sin21.8 = (250*9.81*cos21.8 - 515*sin21.8)*μ
μ = (515*cos21.8 + 250*9.81*sin21.8) / (250*9.81*cos21.8 - 515*sin21.8)
μ = 1388.94/2085.85
μ = 0.6658
Regards
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Ok the first thing u do in solving this problem, is determine what force would be required to move the box across the incline. To do this we need to use the equation:
Cos (21.8) / force across the horizontal .... Using this equation gives us the magnitude of the force going across the incline.
Cos (21.8) / 515 = 555 N
So now we know what force is being applied on the incline. Now for the next part we know that the max static friction force is equal to the force required to get the crate moving, which is 555 N.
The equation for the force of static friction is F = (static fric coefficient)mg
Our only unknown if the coefficient so we arrange the equation to isolate that variable.
Coefficient = F/ mg = 555 N / (250)(9.8) = 555 N / 2450 = .22
Cos (21.8) / force across the horizontal .... Using this equation gives us the magnitude of the force going across the incline.
Cos (21.8) / 515 = 555 N
So now we know what force is being applied on the incline. Now for the next part we know that the max static friction force is equal to the force required to get the crate moving, which is 555 N.
The equation for the force of static friction is F = (static fric coefficient)mg
Our only unknown if the coefficient so we arrange the equation to isolate that variable.
Coefficient = F/ mg = 555 N / (250)(9.8) = 555 N / 2450 = .22