1) To keep a large crate moving across a rough floor, you push down on it at an
ID: 1328363 • Letter: 1
Question
1) To keep a large crate moving across a rough floor, you push down on it at an angle = 23°, as shown in the figure. Find the force necessary to keep the crate moving at constant velocity, given that the mass of the crate is m = 36.0 kg and the coefficient of kinetic friction between the crate and the floor is 0.30. and
2)The system shown in the following figure is in static equilibrium. Given that the mass (m1) is 8.00 kg and the coefficient of static friction between mass (m1) and the surface on which it rests is 0.32, what is the maximum mass that (m2) can have for which the system will still remain in equilibrium?
Explanation / Answer
Let the force be F
Horizontal Component of Force = F *cos(23)
Vertical Component of Force =F *sin(23)
Now, F = m*a
F cos(23) - u* N = m*a
As the Crate is moving with constant velocity , acceleration = 0
Where, N = mg + Fsin(23)
Therefore,
F cos(23) = 0.3 * (36 *g + F sin(23))
F cos(23) - 0.3*Fsin(23) = 0.3 *36g
F (cos(23) - 0.3 sin(23)) = 105.84
F = 131.75N
Force necessary to keep the crate moving at constant velocity,F = 131.75N
As Figure is missing, i have assumed Force to be acting at = 23 below the Horizontal which is generally the case.
Please let me know in case of if there is any problem, with solution.
Please post seperate question in seperate posts only.
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