{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Problem Set 3 Solutions

# Problem Set 3 Solutions - Problem Set III Solutions 1 The...

This preview shows pages 1–4. Sign up to view the full content.

Problem Set III Solutions 1. The block is at rest which means that F x = F y = 0. From Figure 1, it is clear that F x = ma x = 0 = T 1 cos a = T 2 cos b (1) F y = ma y = 0 = T 1 sin a + T 2 sin b = mg. (2) Solving Equation 1 for T 2 and then plugging back into Equation 2 to solve for T 1 , T 2 = T 1 cos a cos b T 1 sin a + cos a cos b sin b = mg = T 1 = mg sin a + cos a tan b . Plugging in numbers: T 1 = (10 kg )(9 . 8 m/s 2 ) 1 2 + 3 2 3 = 49 N T 2 = T 1 3 2 2 = 49 3 N = 85 N 2. (i) For the case where there is no friction between the block and the table, the force on the two block combi- nation is F = ( M + m ) a = - kA = a = - kA M + m . Assuming the block of mass m does not slip, both blocks accelerate at the same rate. The force that is accelerating the smaller block is the force due to friction. Intuitively, if there were no friction between the blocks, the smaller mass would stay at the stretched position and not be pulled back by the spring when the mass M was released. (It would then, of course, fall to the ground.) Therefore, in this case, friction must act in the direction of the motion in order to keep the smaller block on top of the larger block. ma = F f μmg . Since we want to know the maximum distance the block can be pulled, we want the maximum value for the friction. Using the value for acceleration that we found above, ma = m kA m + M μmg A max = μg ( m + M ) k FIG. 1: Problem 1. FIG. 2: Problem 4

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2 (ii) When you add friction between the block and the table, the initial equation for the acceleration changes to F = ( M + m ) a = - kA + μ ( m + M ) g = a = - kA + μ ( m + M ) g M + m . Now the force you need to keep the little block from slipping is ma = m M + m ( kA - μ ( m + M ) g ) μmg A max = ( M + m ) g ( μ + μ ) k 3. The average force is the change in energy divided by the change in distance. Assuming the seat belt keeps the passenger safely in the seat, ¯ F = Δ( E ) Δ( x ) = 1 x f - x 0 1 2 mv 2 f - 1 2 mv 2 0 = 1 0 . 94 m - 1 2 (75 kg )(70 km/h ) 2 = 1 0 . 94 m - 1 2 (75 kg ) 70 km h 1000 m 1 km 1 h 3600 s 2 = - 15 kN 4. There are two forces acting on the mass, the force due to gravity and the force from the spring (see Figure 2). Writing down the forces on the mass, F = ma = kx - mg = x = m ( a + g ) k . The x I just solved for is the distance the spring stretches due to gravity and the acceleration of the elevator, but does not include it’s unstretched length. Therefore, the total length of the spring is given by X = 80 cm + m ( a + g ) k . (3) (a) Plugging into Equation 3 X = 80 cm + (7 . 2 kg )(0 . 95 + 9 . 8)( m/s 2 ) 150 N/m = 132 cm (b) In this case, a=0, so X = 80 cm + (7 . 2 kg )(9 . 8 m/s 2 ) 150 N/m = 127 cm (c) We need to calculate the acceleration and plug it into Equation 3. ¯ a = Δ( v ) Δ( t ) = 0 - 14 m/s 9 . 0 s X = 80 cm + (7 . 2 kg )( - 14 9 + 9 . 8)( m/s 2 ) 150 N/m = 120 cm (d) In this case, we are given X and need to solve for a . X = 3 . 2 m = 0 . 80 m + (7 . 2 kg )( a + 9 . 8 m/s 2 ) 150 N/m = a 40 . 2 m/s 2 To keep the mass from hitting the floor, a cannot be greater than 40 . 2 m/s 2 . (This is assuming that the spring continues to obey the same force law when stretched to this length.)
3 FIG. 3: Problem 5 FIG. 4: Problem 6 5. The forces acting on the mass are the force due to gravity and the tension in the rope, see Figure 3. The mass is accelerating in the x

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern