Question

Can someone help with these

Answer 1

Six

The Formula for the frictional force = mu * N. There is a Resistance offered by the box in contact with the table top. The fact that there is no horizontal force there only means that the box will not move. The frictional force is still there.

Edit

If you look at the comments for this question, you will see that Jason and I have had a very long discussion on the nature of existence. I think he is right about the expected answer, but I'm not sure that the expected answer is the right one. However I will defer my misgivings. Take D (0) as the correct answer. I have not found anyone who would say it is 16 (as I did).

Ff = mu * N

mu = 0.4;

N = 40 N

Ff = 0.4 * 40 N

Ff = 16 N

The kinetic coefficient of friction has nothing to do with the problem.

Seven

Physics is the science of deception and hiding in the corners. This time the number you do not use is the 208 N. Static Friction must be overcome first and that is the larger force of 313 N. The 25 cm/s also has nothing to do with anything.

Remember that weight is a Force, it is not a mass. The only two numbers you need are Ff = N * mu

N = 45 kg * 9.81       Find the normal Force

N = 441.5            

Ff = 313 N                  Given

Ff = N * mu                Formula

313 = 441.5 * mu        Substitute

mu = 313/441.5          Divide by 441.5

mu = 0.7090              Coefficient of static friction

Eight

You do need a diagram for this one. I'll try and do this without one. Draw a ramp. Draw a truck going up the ramp. Connect the crate and the truck with a weightless rope. Put T on the rope. The pointy end of the direction goes up the rope towards the truck. The angle between the ramp and the rope is  33.

There are two force pulling down the ramp: The frictional force. The frictional force ALWAYS GOES AGAINST THE DIRECTION OF MOTION.

The other force is the block's inclination to slide backwards toward the bottom of the ramp. The crate is already a force. 375 N. This force always goes down the Ramp.

So you have 3 forces and for all intents and purposes, they = 0. There is no net force as such.

You can google Mass going up a ramp to see the second force.

F_ramp + Ff = T

The diagrams you look up will tell you that the Ramp Force is

Framp = Sin(theta) * mg

Framp is the force going down the ramp.

mg is the weight of the crate. No conversions needed.

Theta = 33o

F_ramp = 375 * Sin(33)

F_ramp = 204.2

The friction force is the same old formula. Normal * mu

There is a twist. Find those diagrams again.

Ff = Normal * mu

Normal = W * cos(theta)

Normal = 375 * cos(33)  This force is always cos(something as in the incline with the horizontal).

Normal = 314.5

Ff = 314.5 * 0.25

Ff = 78.62 N

So you now have two forces

1. The friction Force which is 78.62

2. The Framp = 204.2

T = Ff + Framp

T = 204.2 + 78.62

T = 283 N

Looks like D to me.

Answer 2

6.  0N.  This questions requires understanding of how friction functions.  Friction is a resistive force, meaning it opposes the direction of any applied or unbalanced forces.  The box in the question experiences no horizontal force, so there is no resistive force in response to it, making it 0N.

7.  This question tests your understanding of static friction.  Static friction only applies when an applied or unbalanced force is applied to an object which does not move.  The static friction always equals the magnitude of the applied or unbalanced force component parallel to the surface which the object rests on.  The question states that the crate starts to move only when the applied force exceeds 313N, so we use this value to determine the force of static friction.  The additional info in the question pertaining to when the crate is moving is irrelevant when determining static friction (only relevant if determining kinetic friction).  Knowing this we solve for the weight of the crate:

F = mg

F = (45)(9.8)

F = 441N = Normal Force

The weight of the crate is also equal to the Normal Force since the object rests on a horizontal surface and the applied force is horizontal as well.  In this question, since the object is not moving at 313N of applied force, the magnitude of static friction equals the applied force:

Ff = μs * Fn

(313) = μs (441)

0.71 (rounded) = μs