The similarities between Chapter 14 and Chapter 18 are both apply concepts of work and energy to solve the problems involving force, velocity and displacement.
Work for variable force:
For particles: U_(1-2)= ∫_(r_1)^(r_2)▒〖F.dr〗=∫_(s_1)^(s_2)▒〖F cos〖θ ds〗 〗
For rigid body: U_F= ∫▒〖F.dr〗= ∫_s▒〖F cos〖θ ds〗 〗
Work of a constant force moving along a straight line:
For particle: U_(1-2)= F_c cosθ(s_2-s_1 )
*the work of F_c represents the area of the rectangle
For rigid body: U_(F_c )=(F_c cosθ)s
*constant magnitude, F_c and constant direction, θ are maintain and the body undergoes a translation, s
Work of a weight:
For particle: U_(1-2)= -WΔy
For rigid body: U_W= -WΔy
*weight, W is negative since W is acting downward and displacement is upwards
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So the equation of motion of tangential direction is ∑▒F_t =ma_t. When kinematic equation a_t=v dv/ds and then integrating both sides,
∑▒〖∫_(s_1)^(s_2)▒〖F_t ds〗=∫_(v_1)^(v_2)▒〖mv dv〗〗
∑▒〖∫_(s_1)^(s_2)▒〖F_t=〗 1/2〗 mv_2^2-1/2 mv_1^2
∑▒〖U_(1-2)=1/2〗 mv_2^2-1/2 mv_1^2 the term on left is the sum of the work done by all forces acting on the particle as the particle moves from point 1 to point 2. The term of T is forms as T=1/2 mv^2 which define the particle’s final and initial kinetic energy.
T_1+∑▒U_(1-2) =T_2
For rigid body: this equation states that the body’s initial and final translational and rotational kinetic energy, plus the work done by all external forces and couple moments acting on the body. Since the body is rigid, no relative movement between these forces occurs, so no internal work is done.
T_1+∑▒U_(1-2) =T_2
Conservative Forces and Potential Energy
Conservative forces is independent of the path and depends only on initial and final positions of the body and particle.
Gravitational Potential Energy
V_g=W_y
If particle is located distance y above an arbitrarily selected datum, the particle’s weight W has positive gravitational potential energy,V_g. If below the datum, V_g is negative since the weight does negative work and at the datum
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Chapter 18 need to develop a means of obtaining the body’s kinetic energy the body is subjected to translation, rotation about a fixed axis or general plain motion.
Chapter 14 has Work of Friction Caused by Sliding which the equation is:
1/2 mv^2+Ps-μ_k Ns=1/2 mv^2
These problems involve cases where a body slides over the surface of another body in the presence of friction. When a block is translating a distance s over a rough surface, the applied force, P just balances the resultant frictional force μ_k N, and equilibrium a constant velocity, v is maintained.
Chapter 18 has the Work of Couple Moment which the couple moment M=Fr. If the body undergoes a differential displacement, then the work done by the couple forces can be found by considering the displacement as the sum of a separate translation plus rotation. When the body translates, the work of each force is produced only by the component of displacement along line of action along the forces.
When the body rotates in the plane through a finite angle ϴ measured in radians, from ϴ_1 to ϴ_2, the work of couple moment