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A 50.0-kg box is resting on a horizontal floor. A force of 250 N directed at an angle of 20.7° below the horizontal is applied to the box. The coefficient of kinetic friction between the box and the surface is 0.300. What is the acceleration of the box?
The answer in the textbook is 1.21m/s^2 but it doesn't show me the working. Pleas explain.
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>>363323
Can you draw the situation for me?
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a 50kg box weights 50*9.8 newtons
that 250n force has a y component of 250sin20.7 and an x component of 250cos20.7
the force on the box by the ground is going to be that weight minus the y component of the force, so 50*9.8 - 250sin20.7
then multiply this bit by 0.300 to find the force of friction in the -x direction
then you can calculate total force in the x direction from 250cos20.7 - that friction number, and then simply apply F=ma
im too lazy to find a calculator and do any of the arithmetic but im sure you can manage that on your own
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>>363328
There is no diagram so I had to make it in paint
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You need to divide the force of 250 N into two parts, one perpendicular to the surface (normal force), the other parallel to the surface (applied force). You use the angle for this.
Now consider the object. The mass applies a force of mass times g increasing the force applied perpendicular to the surface (since the force is applied from below). Let this total force be Fn.
The friction force is Fn times the coefficient 0.3 and reduces the applied force parallel to the surface.
Now you use the formula (applied) Force = mass times acceleration to calculate the acceleration.
(This force must be around 60 N since the mass is 50 kg).
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>>363341
I assume you just ignore the effect the vertical component of the force is going to have on the friction coefficient?
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>>363345
>increasing the force applied perpendicular
you need to add the two forces to obtain Fn
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>>363335
233.86 N being applied in the direction of motion.
.3 * (50*9.8 - (250sin(20.7)) N being applied against the direction of motion.
Net force: 233.86 - 120.48 = 113.37 N
Net accel: 113.37 / 50 = 2.26 m/s^2
2 possibilities here:
1) The book's wrong, either in its solution or writing the problem.
2) You're pulling at an angle 20.7deg below the horizontal, not pushing.
In either case that means we add the vertical force component to the box instead of subtracting it, and the math works out to 1.207 m/s^2 instead.
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