from your work I guess you are pulling at 36.87 deg up and not pushing down. If so I agree with the 4.84 m/s^2
now the sum of forces on the upper package must give an acceleration of 4.84 m/s^2
20 (4.84) = 400 cos 36.87 - [200 -400 sin 36.87] mu
hmmm, you must be pulling up on the lower package, because 400 sin 36.87 would lift the upper one right off. In that case
20*4.84 = 200 mu
mu = .484
There are two things you did not tell me so I had to assume.
1. You are pulling up, not pushing down.
2. You must be pulling up on the lower package or the top one would lift off.
If these assumptions are incorrect, you must change signs all over the place.
A person must move two heavy packages over a surface level so they place a 20kg package over a 30kg package and decide to move them together. The coefficient of kinetic friction between the lower package and the ground is 0.3. The person exerts a force of 400N at an angle of 36.87degrees from the horizontal.
Assuming that the upper package doesn't slide off the lower, determine the rate at which the accelerate.
I found the acceleration
Fn + Faperp - Fg = 0
Fn + 400sin36.87-500 = 0
Fn = 260N
Fa - Ffr = ma
400cos36.87 - 0.3*260 = 50a
a = 4.84m/s^2
Now how do I determine the minimum coefficient of static friction that must exist between the objects so as to ensure that the upper package does not slide of the lower one.
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The coefficient of kinetic friction between the surface and the objects resting on it is 0.2. There are three objects, one weighing 25kg and connected by a rope to an object weighing 35kg which is connected by a rope to an object weighing 60kg. Object 1 and 2 are on a ramp that is 36.87 degrees to the horizontal and object 3 is hanging off the edge by a pulley system. Determine the acceleration of the objects and the tension in each rope.
I created 3 equations for acceleration
Ft1 - 40 - 150 = 25a
Ft2 - Ft1 - 56 - 210 = 35a
600 - Ft2 = 60a
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a = 1.87
FT1 = 236.75N
FT2 = 487.8N
5 answers
In the diagram the person is pushing down on the packages. What would change?
Friction changes. The vertical componentof the pushing force adds to weight, causing the force of friction to increase.
If up was positive, would the vertical component of the pushing force be negative?
No.
0 answers