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You need to review the trigonometric functions. In your question, assume the forces are to be resolved into component forces along the orthogonal axes r & q, $\sum F_r = Rcos\theta - Ncos\theta = (R-N)cos\theta$ $\sum F_q = F + Rsin\theta - Nsin\theta = F + (R-N)sin\theta$ If you have any doubt, you can prove the validity of the component forces by the ...


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In order to make the notation shorter for the remaining calculations in my answer I will denote the matrix with $$ H = \frac{1}{2} \begin{bmatrix} -q^\top \\ q_0\,I_3 + q^\times \end{bmatrix}. $$ In this case you have four equations but only three unknowns. So there might not be a solution that exactly matches all equations, especially when the time ...


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As shown in this thread, the acceleration of a mass on an incline equals $g*sin\theta$. When the inclination is lowered with a deacceleration of -$a_{dec}$, from relativity, it means the mass is moving backward from the original frame, thus, $a = (g - (-a_{dec}))sin\theta = (g + a_{dec})sin\theta$.


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The diagrams are the following On the mass (lets call it A) act only the weight and the reaction from the inclined floor ($N_1$). Also, from the kinetic diagram you see the accelerations. As you can see on the inertial Frame you have: $a_A$ the acceleration of A with respect to the inertial frame. (the direction is not known) $a_E$ is the acceleration of ...


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If you are asking why does the 1 kg mass (I'll denote it as $m_1$), moves towards the inclined plane, then IMHO the exercise you shouldn't have any friction. If there is no friction, then the $m_1$ will move downwards, because there will be no horizontal force to keep it attached to the Wedge (I suspect its mass is 5 kg, so I will denote it as $m_5$. Below ...


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Let's assume for now the wedge S massless. its mass is not going the change the aspect of the behavior of the One kg mass. The components of the 1 kg along and perpendicular to the ramp are $$F_{parallel}=mgsin37 =1*9.8*0.601 \quad and \quad F_{perp} =1*9.8cos37$$ The acceleration of the wedge S is $$\alpha =F_{parallel}/1=gsin37=9.8*0.601=5.89m/s^2$$ And ...


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Ok, will answer both of your questions here, you may want to merge them into one question. The reason you are having problems is that you have heavily overloaded the word pair in your source document. So everything they are talking about is a pair. However, not all instances of word pair means the same thing. So while the chapter previous to this chapter ...


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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, this can be expressed as $a = (F_2 - F_1)/m = (130 - 30)/10 = 10 m/s^2$. Now we can find the final velocity, $V_F = V_0 + a*t = 10 + 10*2 = 30 m/s^2$. For $V_0 = ...


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If I understand the problem the you have: a mass moving at 10[m/s] at time t=0, a force of 30 [N] is applied, and by time t=2 it increases to 130 [N] The question is what is the final momentum. The final momentum (at time $t_2=2[sec]$) will be equal to the initial momentum plus the impulse of the force (see below): $$m\cdot v_2 = m\cdot v_0 + \int_{t_0}^{...


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There is no need to know the mass of the block. If it were movable yes, but in this case you can assume that is a fixed wall. So the whole idea is that you are assuming a constant (average) deceleration of the bullet. Since you already know the deceleration on the bullet, then from the equation $$\sum F = m\cdot a$$ you know that the only force on the X axis ...


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Planetary gear? Both shafts will rotate at a different speed / transmit a different torque, depending on what you're after. Electric drills often use these gearboxes if they have multiple speeds.


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No. You cannot increase both torque and speed without adding more energy to a system. In a gear train/belt & pulley combo, you have one member rotating slower but with higher torque and the other faster but with lower torque. You would need an outside power source to further increase the speed or torque beyond what you input, per the conservation of ...


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A simple machine is what you're referring to. A simple machine turns motion into mechanical advantage. Examples include: lever inclined plane pulleys wheel & axle gears inclined plane/wedge/screw These machines convert extra motion or rotation into mechanical advantage so that the same work can be done with less applied force.


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