diagram:
big triangle mass m with the square angle bottom left and angle theta bottom right.no frition with the horizontal surface. Above that triangle a similar triangle mass m with a shorter hypothenuse (square angle top right, angle theta top left, no friction between both triangle).On top of these triangles,a block mass m ,with friction between the block and the top triangle.
In the diagram, both the prisms and the block have equal masses m. Angle θ is given. Both surfaces of the larger prism are frictionless; however, there is friction between the horizontal surface of the smaller prism and the block. A horizontal and constant force of unknown magnitude F is exerted on the larger prism. As a result, the three objects remain at rest relative to each other.
(a) Find the magnitude of the acceleration of the larger prism a.
1)a=2gtanθ
2)a=gtanθ
3)a=gsinθ
4)a=2gcosθ
5)a=gcosθ
6)a=2gsinθ
(b) Find the value of the pushing force F.
1)F=6mgtanθ
2)F=6mgsinθ
3)F=6mgcosθ
4)F=3mgcosθ
5)F=3mgsinθ
6)F=3mgtanθ
(c) Find the minimum coefficient of static friction μs between the block and the smaller prism that makes it possible for the block to stay at rest relative to the prism.
1)μs=2tanθ
2)μs=tanθ
3)μs=2sinθ
4)μs=sinθ
5)μs=2cosθ
6)μs=cosθ
the coeff of friction min...I donc think it has to do with the tan, since µ has to be at most equal to 1.
c) µs = tan theta
a) a=g*tan(theta)
b) F=3*m*g*tan(theta)
Thank you very much guys
To solve this problem, we need to analyze the forces acting on each object and apply Newton's laws of motion.
For the larger prism:
(a) The only force acting on the larger prism horizontally is the pushing force F. There is no friction between the larger prism and the horizontal surface, so there is no opposing force in the horizontal direction. Hence, the acceleration of the larger prism is zero, so the correct answer is (a) a = 2gtanθ.
For the block on top of the smaller prism:
(b) The frictional force acts horizontally in the opposite direction to the pushing force F. Since the block remains at rest relative to the smaller prism, the frictional force must balance the pushing force. Therefore, the correct answer is (b) F = 6mgsinθ.
For the smaller prism:
Since the block is at rest relative to the smaller prism, there must be enough friction between them to prevent the block from sliding. This gives us the minimum coefficient of static friction (μs) between the block and the smaller prism.
(c) To find μs, we need to consider the forces acting on the block in the vertical direction. The weight of the block is mg and the normal force exerted by the smaller prism is also mg. The maximum frictional force that can be exerted between two surfaces in contact is given by μs times the normal force (mg).
Therefore, the maximum frictional force is μs * mg. To keep the block at rest, the frictional force must be greater than or equal to the force trying to slide the block (F). So, the inequality μs * mg ≥ F should hold.
Since F = 6mgsinθ, we can substitute this in the inequality to get:
μs * mg ≥ 6mgsinθ
Dividing both sides by mg, we get:
μs ≥ 6sinθ
Since the coefficient of friction cannot exceed 1, the minimum value of μs will be the maximum value obtained from the inequality. Therefore, the correct answer is (d) μs = sinθ.
In summary, the correct answers are:
(a) a = 2gtanθ
(b) F = 6mgsinθ
(c) μs = sinθ