4 homework problems (static coefficient, acceleration/direction, power, ect)

[jolin4bj]

#1) The coefficient of static friction bwtween the block and the board is 0.6. If the blocks weight is 75 N, what minumum weight attached to the pulley will cause the block to move?

#2)Find the direction and acceleration of a single pulley with 2 masses (7 kg and 10 kg)

#3)A machine requires 30J of energy every minute what is the power consumption?

#4)A car moving north at 60 km/h collides and lockes bumpers with a large trunk going 50 km/h West. If the mass of the car is only 1/3 that of the truck mass, what is the magnitude and direction of velocity of the combined masses?

1~ I don't even know where to start on #1...

2~ I think you start with the force equation F= ma

3~ I know the power equation is P= W/t, but how do you get work when all we are given is 30 J

4~ I know the direction is northwest, but I'm not sure how to find the magnitude of the combined masses??

 

[Arman_AAT_Khos]

1.) My guess is; this is the situation: one block is resting on the board some horizontal distance from the pulley which has the other block hanging vertically from it.

F = F_f : impending motion
Belt Friction equation for a pulley system is: F_tight = F_loose * e ^ (u * theta)
F_tight is the force on the taut rope hanging vertically, or intuitively the side with the least slack.
F_loose is the force on the rope that has slack, that is, the side where the rope has the least amount of surface contact with the pulley.
theta (in radians) is the wrap angle defined as the angle between the contact of the tight and loose sides of the rope on the pulley. So if one rope hangs vertically and the other is horizontal, theta is pi/2. If they are both vertical then theta is pi.

2.) The Force on the pulley is equivalent to the sum of the tensions. T_pulley = T_1 + T_2. Setup equilibruim equations for the tensions.

3.) W = delta(E) = E_2 - E_1 pick sample points and times plug into P = dW/dt

4.) Draw out velocities as vectors, and use conservation of momentum to find the final velocity of the combined truck and car mass.- Arman Khos.

 

[quicknote]

I can provide some guidance and solutions for the homework problems. Please note that the solutions provided here are for educational purposes and may not be the exact answers to the problems given in the homework.

#1) The coefficient of static friction is a measure of the force needed to overcome the friction between two surfaces and cause motion. In this problem, we are given the coefficient of static friction (μs) as 0.6 and the weight of the block (W) as 75 N. To calculate the minimum weight (F) needed to cause the block to move, we can use the formula F = μs * W.

F = 0.6 * 75 N = 45 N

Therefore, a minimum weight of 45 N attached to the pulley will cause the block to move.

#2) To find the direction and acceleration of a single pulley with two masses, we can use Newton's Second Law of Motion, which states that the net force on an object is equal to its mass multiplied by its acceleration (F = ma).

First, we need to calculate the net force acting on the system. This can be done by considering the forces acting on each mass separately. The 7 kg mass has a force of 7 kg * 9.8 m/s^2 = 68.6 N acting downwards due to gravity. The 10 kg mass has a force of 10 kg * 9.8 m/s^2 = 98 N acting downwards due to gravity. Since the pulley is massless, it does not contribute to the net force.

Net force = 98 N - 68.6 N = 29.4 N

Next, we can use the net force and the total mass of the system (17 kg) to calculate the acceleration using the formula a = F/m.

a = 29.4 N / 17 kg = 1.73 m/s^2

Since the net force is acting downwards, the acceleration will also be downwards. Therefore, the direction of acceleration is downwards.

#3) Power is the rate at which work is done or energy is transferred. In this problem, we are given the energy consumption (30 J) and the time (1 minute). To calculate the power, we can use the formula P = W/t.

P = 30 J / 60 s = 0.5 J/s

Therefore, the power consumption is