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$\sum V$ = total force = 118 + 15 = 133 Distance of the total force to the rear wheel ($x$): $\sum M$ about the rear wheel = 0 $x = 15*9/133 = 1.015 m$ - the center of the whole mass is located 1.015 …

Work done is related to motion, which usually results in change of positions. In your picture the rotation of the arms produces positive work, if a counter C_b is added, it means partial work is negat …

It is the effect of the "inertia force", which always occurs during the event of "sudden acceleration, or deacceleration", and always in the reversed direction of the acceleration, or deacceleration. …

No. The equation you wrote represents the change in kinetic energy due to a change in the speed of motion. Energy is required to start the motion, in a vacuum, the body stores the terminal energy with …

The sketch below is not a response to the question, but to show the final positions of the components (all in red), which may be helpful in deriving the correct solution, if any. Note that it is assum …

I see this problem differently. The 10kg mass was moving at a constant speed at 10 m/s, so $V_0 = 10 m/s$. 2 seconds After a 30N force was applied, the force increased to 130N due to acceleration, thi …

The figure below shall answer your question. Moving from point B to C, the sliding force $F_S = mgsin\theta = ma$, since no friction, $a = gsin\theta$, and $\theta = Tan^{-1}(L_{BC}/L_{AC})$

Isn't this a math problem that can be arranged as, $T cos 45 = mg ---> \frac{T cos 45}{cos 45} = \frac{mg}{cos 45} ---> T = mg/cos 45$ There are two accelerations - centripetal (towards the center) …

The speed of rotation has two components - rotational speed (rotation/time) and linear (tangential) speed (distance/time). As shown on left in the sketches below, the rotation speed is constant about …

Constraint Method for Pulley Problems Assumptions: The string is taut and inextensible at each and every point of time. The string is massless and hence the tension is uniform throughout. Pulley is …

Revised (after @DKNguyen's comment) The force diagram below contains both cases, which should result in the same conclusion for the mass of block A. (For both cases, the only effect left from the bloc …

This is a simple static problem, with the assumption that all contact surfaces are frictionless. In the static position (no motion), the cable has a tension equal to 2g. Then, in order to move the 2kg …

No friction Two outcomes after the initial push: a) run out on the curve, and b) slide down the slope. With friction The outward centrifugal force is overcome by the inward friction force $\mu N o …

$a_t = \frac{dv}{dt} = \frac{d(r∗ω)}{dt} = r* α$. Here, $v$ is a vector. $a_c = r*ω^2 = \frac{v^2}{r}$. Here $v$ is a scaler. In your numerical example, both changes in velocity and time are given, so …

You need to find the compressibility factor, which is defined as $Z = \dfrac{pV}{nRT}$ of the hydrogen, then correlate to the "ideal gas", which has a compressibility factor of 1. https://en.wikipedia …