a. Two strips of metal are pinned together as shown, with a rod of 10mmdiameter. The Ultimate shear stress for the rod is 60MPa. Determine themaximum force required to break the pin.EV (6)b)Two students are watching an action film in which a car drives down a ramp onto the back of amoving lorry. Both are moving at high speed, the car slightly faster than the lory.One student complains that this is impossible because the car would not be able to stop beforehitting the cab of the lorry.The car has mass 1250 kg and is moving at a speed of 28.0 ms. The lorry has mass 3500 kgand a speed of 25.5 ms. The length of the flat back of the lorry allows a braking distance of5.0 m.By considering both momentum and energy show that the stunt is possible, provided aminimum force of about 600 N slows the car down. You should support your explanations withcalculations.Treat the situation as one in which two objects join together.CR (8)7c.Explain Newtons Second Law of motion in terms of momentum.AN (6)

Answered: Image /qna-images/answer/5398e696-1126-43b5-9ff8-98e225193f2d.jpg

Answered: a. Two strips of metal are pinned…

a. Two strips of metal are pinned together as shown, with a rod of 10mm diameter. The Ultimate shear stress for the rod is 60MPA. Determine the maximum force required to break the pin. F F

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Related questions

Q: 0.25 in. 5 in. 1 in. 0.25 in. -0.25 in. 2.5 in.

A: Maximum axial force P that can be applied to the steel plate

Q: The wood beam has an allowable shear stress of Tallow = 5 MPa. (Figure 1) Figure 50 mm 50 mm 50 mm…

Q: A 100 m long cable with a cross-sectional area of 1.0 mm² stretches by 5.0 cm when a force of 1,000…

A: Given data, Length of cable l = 100 m. Cross sectional area A=1 mm2=10-6 m2. Amount of cable…

Q: Assume if the shear stress in steel exceeds about 4.00 x 108 N/m² the steel ruptures. (a) Determine…

A: The shear stress in steel is, S=4.0×108 N/m2 The diameter of the steel bolt is, D=1.25 cm The…

Q: , determine the displacement of point B.

Q: What hanging mass will stretch a 2.0-mm-long, 0.59-mmmm-diameter steel wire by 1.2 mmmm ? The…

Q: Problem 2. From the given section of a wide flange shown in the figure, it carries a vertical shear…

Q: A heavy crate with weight 4900 N hangs from a thick nylon rope. The rope has a cross-sectional area…

A: Solution: Given that Weight W = 4900 N Area A = 1.4x10-4 m2 youngs modulus E =3.7x109 N/m2

Q: A metal slab of side (74+ 75) cm and thickness (74 + 25) cm is subjected to a tangential force of…

Q: a) Figure Q2a shows a hollow tube of rectangular cross-section subjected to an axial compressive…

A: Write the given values with the suitable variables. P=250 kN=250×103 Nb=0.12 md=0.05 mt=0.003 m…

Q: A cable used to lift heavy materials like steel I-beams must be strong enough to resist breaking…

Q: A 1.05-m-long rod of negligible weight is supported at its ends by wires A and B of equal length…

A: Solution:-

Q: P4. If the cross section of the femur at section a-a can be approximated as a hollow as shown,…

A: Maximum normal stress = 235.98 lb/in2 = 236 psi Detailed solution is given below: Please UPVOTE sir

Q: A force is applied to a circular surface with area 70 cm². The surface is parallel to the x-y plane.…

Q: NO 4. End A of the bar AB in Fig. P11.53 rests on a fric- tionless horizontal surface, and end B is…

A: Given:- We have given a figure as the end of bar AB is in rest on a frictionless horizontal…

Q: The truss is composed of equilateral triangles of side a-1.9 m. The load is L=310 N. Determine the…

Q: For the beam and loading shown, determine the maximum absolute value of bending moment and shear.

A: Draw the free-body diagram of the beam.

Q: A steel wire of length 2.10 m with circular cross section must stretch no more than 0.300 cm when a…

Q: Problem 6: The assembly shown below consists of 30 mm diameter aluminum bar ABC with fixed collar at…

Concept explainers

Mechanical Properties of Solids

There are three types of matter- solid, liquid and gas. Out of them, solids have a rigid shaped and size. The intermolecular forces of attractions in solids are greatest. Mechanical properties are the physical properties of a material which it shows, when it is provided with certain amount of force or load. These properties are to be kept in mind for various purposes. When a force is applied to an object, the object may or may not get deformed depending upon its elastic property. But every time the object develops a restoring force against the force applied. This is commonly known as stress. If the object changes its dimensions due to the force applied it develops strain ( which is the ratio of change in dimension to the original dimension).

Question

a. Two strips of metal are pinned together as shown, with a rod of 10mm
diameter. The Ultimate shear stress for the rod is 60MPa. Determine the
maximum force required to break the pin.
EV (6)
b)
Two students are watching an action film in which a car drives down a ramp onto the back of a
moving lorry. Both are moving at high speed, the car slightly faster than the lory.
One student complains that this is impossible because the car would not be able to stop before
hitting the cab of the lorry.
The car has mass 1250 kg and is moving at a speed of 28.0 ms. The lorry has mass 3500 kg
and a speed of 25.5 ms. The length of the flat back of the lorry allows a braking distance of
5.0 m.
By considering both momentum and energy show that the stunt is possible, provided a
minimum force of about 600 N slows the car down. You should support your explanations with
calculations.
Treat the situation as one in which two objects join together.
CR (8)
7
c.Explain Newtons Second Law of motion in terms of momentum.
AN (6)

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step by step

Solved in 2 steps with 2 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions

Block A is held in equilibrium when force F = 140 lbf is applied. Each cable is made out of 5-0s gage B & S Non Ferrous Metals wire. Determine the normal stress in each cable and the mass of block A. You may assume the cables and pulleys have negligible mass. C Be A D E F
A man with a mass of 72.1 kg stands on one foot. His femur has a cross-sectional area of 8.00 cm2 and an uncompressed length 49.6 cm. Young’s modulus for compression of the human femur is 9.40 × 109 N/m2. A.) How much shorter is the femur when the man stands on one foot? B.)What is the fractional length change of the femur when the person moves from standing on two feet to standing on one foot?
2. A Torque of magnitude T = 900 N.m is applied at D as shown. Knowing that the diameter of shaft AB is is 60 mm and that the diameter of shaft CD is 45mm, determine the maximum shearing stress in (a) shaft AB, (b) shaft CD. A որոշու B 40 mm 100 mm Figure 2 T = 900 N · m
A force stretches a wire by 1.0 mm.a. A second wire of the same material has the same cross section and twice the length. How far will it be stretched by the same force?b. A third wire of the same material has the same length and twice the diameter as the first. How far will it be stretched by the same force?
Assume that if the shear stress in steel exceeds about 4.00 x 108 N/m², the steel ruptures. (a) Determine the shearing force necessary to shear a steel bolt 1.40 cm in diameter. IN (b) Determine the shearing force necessary to punch a 1.25-cm-diameter hole in a steel plate 0.550 cm thick. N
Structural member AB is to be supported by a strut CD. Determine the smallest length CD may have, and specify where D must be located for a strut of this length to be used. Take x= 120 mm. 60 mm D. y 20 mm E The smallest length CD may have is mm. The coordinates of point Dare ( and Imm.
Problem 2 P A 40 kips 30 in. 30 kips B 30 kips 40 in. Two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Determine the magnitude of the force P for which the tensile stress in rod AB is twice the magnitude of the compressive stress in rod BC.