I am study for an exam on Monday and I am trying to figure out these questions,

Question

I am study for an exam on Monday and I am trying to figure out these questions, they are show your work questions:

1. A jet lands on aircraft carrier at 180mi/hr

a.) What is its acceleration (assume its constant) in SI units if it stops in 2.5 due to an arresting cable that snags the airplane and brings it to a stop?

b.) If the airplane touches down at positions x=0, what is its positions 1.5s after?

2.

A 1520-N crate is to be held in place on a ramp

that rises at 30.0° above the horizontal (see

figure below). The massless rope attached to the crate

makes a 22.0° angle above the surface of the

ramp. The coefficients of friction between the

crate and the surface of the ramp are ?k = 0.450

and ?s = 0.650. The pulley has no appreciable

mass or friction. Assume the weight hanging

from the pulley is the MAXIMUM that can be

used to hold the crate stationary on the ramp.

a) Draw a FBD for the relevant bodies.

To avoid confusion use se Fg for the weight of the crate

and w for the weight of the hanging mass.

b) Write the equations for the FBD in part (a) by applying Newton’s 2nd Law.

c) What is the weight of the hanging object?

Explanation / Answer

2.

Weight of the crate Wc = 1520N
Angle of ramp to horizontal A = 30deg
Coefficient of static friction of crate s= 0.650

Angle of rope attachment to ramp B = 22deg

Stage 1: Before the rope is attached
---------------------------------------...
The weight of the crate normal to the ramp is given by:

(1) Wcn = Wc * cos(30) = 1520 * 0.866 = 1316N

The static friction force, which by definition is parallel to the ramp, is therefore:

(2) Wcsf = Wcn * s = 1316 * 0.650 = 856N

This is the resistance of the crate to moving in either direction on the ramp.

The force of gravity parallel to the ramp pulling down on the crate is:

(3) Wcgr = Wc * sin(30) = 1520 * .5 = 760N

Because

(4) 856 - 760 = 96N > 0, the crate at rest is prevented from sliding down the ramp, absent other forces.

(5) Wcsf = (Wcn - Wn) * s = (1316 - Wn) * 0.650.

(6) Now, Wr = W * cos(B) = W * cos(22) = W * 0.927

For W to equal gravity plus static friction and thus hold the crate still but not move it:

(7) Wr = Wcgr + Wcsf

Substituting equations (6) for Wr, (3) for Wcgr and (5) for Wcsf, we get:

(8) 0.927W = 760 + 855 - 0.650Wn

However, Wn = W * sin(22) = 0.375W; substituting that in (8) we get:

(9) 0.927W = 1615 - 0.650 * 0.375W = 1615 - 0.2440W

Solving for W:

(10) W = 1615 / 1.17 = 1380N

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