A 111-kg (including all the contents) rocket has an engine that produces a const

Question

A 111-kg (including all the contents) rocket has an engine that produces a constant vertical force (the thrust) of 1730 N . Inside this rocket, a 14.1-N electrical power supply rests on the floor.

(I already found the acceleration which was 5.79)

When it has reached an altitude of 120 m, how hard does the floor push on the power supply?

You are lowering two boxes, one on top of the other, down the ramp shown in the figure (Figure 1) by pulling on a rope parallel to the surface of the ramp. Both boxes move together at a constant speed of 13.0 cm/s . The coefficient of kinetic friction between the ramp and the lower box is 0.416, and the coefficient of static friction between the two boxes is 0.752.

What force do you need to exert to accomplish this?

Explanation / Answer

Question 1:

F = ma

a = F / m

acceleration = (Thrust - mg) / m = (1730 - 111 * 9.81) / 111 = 5.7756 m/s2

For the last part altitude doesn't really matter, it is just used to show you the acceleration due to gravity is still the same and we can't think that we are in outer space where gravity is zero.

Electric power supply mass = 14.1 / 9.81 = m = 1.437 Kg, now add the two accelerations and multiply by mass. we add them because they are in the same direction and both acting upon the power supply.

m * ( a + g ) = 1.437 (5.7756 + 9.81) = 22.3965 N

If we know the weight of it under these circumstances, and according to Newton's third law, we know the normal force is pushing up on it by the exact same force.

Therefore, Normal force is pushing up on the power supply 22.3965 N

Question 2:

First, we have to find the angle of inclination of the ramp. This is the angle the ramp is raised to above the horizon

theta = arctan(2.5 / 4.75) = 27.759 degrees.

Since the two boxes are sitting on top of each other, we can take both of them as a system and we can add their masses.

Now, we have to find all the forces on the boxes that made them to move with uniform speed (which means net force along the inclined plane = 0)

Normal force of the box acting on the ramp = cos27.759 * (32 + 48) * 9.81 = 694.4838 N

Now, force of friction = muk * N = 694.4838 * (0.416) = 288.9053 N

Now, working on finding the forces working that pull the boxes down the ramp, we get the equation

= mg * sin (theta) = sin27.759 * 784.8 = 365.52 N

Fnet = 365.52 - f friction = 365.52 - 288.9053 = 76.6147 N

This is the net force acting along the incline downwards. To move with constant velocity. Net Acceleration must be zero. Which means the above force should be canceled by the Man.

The force that the man should exert =  76.6147 N

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