# What is physically changing from velocity or acceleration to force and their vector components?

Here is question with two pulleys. https://physics.stackexchange.com/questions/342861/why-is-velocity-of-the-mass-v-cos-θ-why-not-2v-cos-θ

Its clear that $$u =\frac{v}{\cos \theta} \tag{1}$$

But now consider same arrangement with at both ends instead of $$v$$ are being pulled by force $$F$$ and the upward force acting on mass will be $$F'$$. Here

$$F'=2F\cos\theta \tag{2}$$

If we consider velocity or acceleration of point $$P$$ compared to that of mass, it shall be given by equation $$(1)$$.

So, my question is, what is changing in both scenarios that we give force on mass is component of force by rope but in both cases velocity of rope is component of velocity of mass? Because mass is a scalar quantity, so why is the nature of force different than that of acceleration (or velocity)?

Consider the energy or power balance of the phenomenon simplified by assuming uniform speeds $$u$$ and $$v$$ during the analyzed motion, neglecting power losses due to the friction and assuming that the mass of the rope and pulley system vanishes. In both cases we have from the geometry and thus kinematics that $$u=\frac{v}{\cos\theta}$$ and from Newton's second law of motion that $$F'=2F\cos\theta$$. Notice that in what follows, we do not explicitly applied Newton's second law of motion to the point mass but recover it from the law of conservation of mechanical energy (or the more dependable statement of the work-energy theorem). Applying the work-energy theorem we have $$\int_0^t F' \cdot u(t) - mg \int_0^t \frac{d}{dt} u(t) \; dt = 2 \int_0^t F \cdot \frac{v(t)}{\cos \theta} - mg \int_0^t u(t) \; dt = \frac{1}{2} m (u(t)^2 - u(0)^2) = 0$$, where the motion occurs over the time period $$[0,t]$$. This equation then recovers the consistent Newton's second law of motion $$(F' - mg) \cdot u = 0$$ or $$F' = mg$$ and $$2F \cdot v - mg \cdot \frac{v}{\cos \theta} = 0$$ or $$F \cos \theta = \frac{mg}{2}$$, as applied to the point mass. An interpretation which expresses this consistency of the mathematical description of the phenomenon is that the factor $$\cos \theta$$ shifts from the force to the speeds to preserve the consistency.
• @UV0 let us assume that in both situations the mass moves at constant velocity. Denoting $v':=u$ for clarity, the relationships $v'=u=\frac{v}{cos\theta}$ is derived by geometry (and then kinematics) and $F'=2F\cos\theta$ derived from Newton's laws are true in both situations depicted in the pictures in the OP. We observe that the factor $\cos\theta$ has flipped sides or gets applied to the tension in the rope in one equation and the velocity of the mass in the other. (1/2) Apr 4, 2021 at 19:12
• @UV0 the idea in the explanation above is proof by contradiction. If we assume, for instance, the wrong kinematic $v'\cos\theta=v$ (instead of the accurate kinematic) and apply the law of conservation of mechanical energy to the point mass system and the $\text{pint mass}\cup\text{rope and pulleys}$ respectively, we obtain $F'=mg$ and $2Fv-mgv'=2Fv-mgv\cos\theta=0$ so that $F=\frac{mg}{2}\cos\theta$ which implies that $F'=2F\cdot\frac{1}{\cos\theta}$. The final expression obtained contradicts Newton's second law $F'=2F\cos\theta$. Therefore the assumption $v'\cos\theta=v$ is incorrect. (2/2) Apr 4, 2021 at 19:43
• The answer above alludes to the proof by contradiction by indicating that the shifting of the $\cos\theta$ factor is what enables consistency in the mathematics of the related principles of geometry $-$ Newton's laws $-$ conservation of energy. In the comments above, we observe that an erroneous geometric, and thus kinematic, relationship between the velocities of the masses and rope lead to a contradiction to the Newton's second law of motion. The only case in which this contradiction is resolved is in the case that $v':=u=\frac{v}{\cos\theta}$. Apr 4, 2021 at 19:49