# engineering science

An object weighing 60N rest on a slope forming an angle of 17.5 degrees with the horizontal. A force of 44N parallel to the sloping surface just moves the object upwards. Calculate the frictional force.

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1. break the 44 N force into components up the slope, and normal to the slope.
F=44cos17.5 u + 44sin17.5 n

now break the weight in to normal, and up the slope components.

60N down= 60cos17.5 n - 60 sin17.5 u

frictional force opposes upward, so it is a negative force up the slope.

Now consider just the n forces.
friction=mu(60cos17.5+44sin17.5)
well, at this point, you are stuck without mu. So,
compute that

then up the slope...
fu=friction
44cos17.5 u-60 sin17.5=friction
now you can compute friction, and go back with the normal equation, and compute mu.

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2. m*g = 60 N. = Wt. of object.

Fp = 60*sin17.5 = 18.04 N. = Force
parallel to the incline.

Fn = 60*cos17.5 = 57.22 N. = Normal =
force perpendicular to the incline.

!@#\$%^&-Fp-Fs = m*a
44-18.04-Fs = m*0 = 0
Fs = 44-18.04 = 26 N. = Force of static
friction.

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posted by Henry

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