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4 years ago

Physics friction question help

Physics

2 answers:

Alborosie4 years ago

8 0

Assuming that acceleration due to gravity is g = 9.8 m/s^2.

mass of block = 5 kg
μ = 0.2
applied force = 9 N

The force of gravity, Fg, on this block is given by Fg = m * g --> F = 5 kg * 9.8m/s^2 = 49 N.
The normal force on this block exerted by the ground is Fn = Fg --> 49 N. (It points in direction opposite to Fg, but we don't need to worry about that here.)
The maximal force of friction on the block that resists its movement is Ff = Fn * μ --> Ff = 49 N * 0.2 = 9.8 N

So, the max frictional force is 9.8 N.
Since the applied force is only 9 N, the block does not move.
Since we deduced that the block is not moving, then the block must be in equilibrium (all external forces must equal zero). So, the frictional force must match the applied force of 9 N.

padilas [110]4 years ago

4 0

Frictional force = (coefficient of friction) x (mg).

a). If the coefficient of static friction is 0.2 and m= 5 kg and g= 9.8 m/s^2, then frictional force= (0.2) x (5 kg x 9.8 m/s^2) = 9.8 N. This is the maximum frictional force that prevents the block from moving.

b). To answer this question, you must first find the Net force. You find this by simply subtracting the force vectors from each other. Because both the frictional force and applied force are horizontal, you can directly subtract. If 9N is applied to move the block but there is a maximum frictional force of 9.8 N, then the block does not move because the applied force cannot overcome the frictional force. 9<9.8 (9 N - 9.8N= -0.8 N). The negative sign indicates that the net force would be in the leftward direction ( the direction the friction force is pointing).

C). The size of the frictional force always matches the size of the applied force if the applied force is less than the frictional force. This is because frictional force cannot move something, it can only prevent movement. However, if the applied force is greater than the frictional force, then the frictional force would be at its largest magnitude. Since the first situation is the case here, the force of friction = applied force = 9 N.

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$F_{AB} = F_{AB} i + F_{AB}j +F_{AB}k\\F_{AC} = F_{AC} i + F_{AC}j +F_{AC}k\\F_{AD} = F_{AD} i + F_{AD}j +F_{AD}k\\W = -160 k$

Equation of Equilibrium:

$Sum of forces = 0\\F_{AB}. u_{AB} + F_{AC}.u_{AC} + F_{AD}.u_{AD} + W = 0\\(-0.857F_{AB}i + 0.429F_{AB}j +0.286F_{AB}k) + (-0.857F_{AC}i - 0.286F_{AC}j +0.429F_{AC}k) + (+1F_{AD} i) + (-160k) = 0$

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$-0.857F_{AB} -0.857F_{AC} +1F_{AD} = 0\\+ 0.429F_{AB} - 0.286F_{AC} = 0\\+0.286F_{AB} +0.429F_{AC} = 160$

Solving Above equation simultaneously we get:

$F_{AB} = 172.1356\\F_{AC} = 258.2033\\F_{AD} = 368.8004$

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