The torque about the origin O due to a force F acting on an object with the position vector r is the vector quantity ? = r × F. If several forces F_j act at positions r_j, then the net torque (units: N·m or lb·ft) is the sum ? = ? r_j × F_j Torque measures how much the force causes the object to rotate. By Newton's Laws, ? is equal to the rate of change of angular momentum. Calculate the net torque about O at P assuming that a 25-kg mass is attached at P, F = 45 N and ? = 30° (see the figure). [Figure: rod of length 10 m from O to P at angle ?; force F applied at P at 125°; weight F_g acting downward] The force F_g due to gravity on a mass m has magnitude 9.8 m/s²·kg in the downward direction. (Use decimal notation. Give your answer to two decimal places.)
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First, we need to find the position vector r. In this case, r is the vector from the origin O to point P. Since P is 10 m away from O and at an angle of 30 degrees, we can find the components of r as follows: r_x = 10 * cos(30) = 10 * (√3 / 2) = 5√3 m r_y = 10 * Show more…
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The torque about the origin $O$ due to a force $\mathbf{F}$ acting on an object with position vector $\mathbf{r}$ is the vector quantity $\mathbf{\tau}=\mathbf{r} \times \mathbf{F}$. If several forces $\mathbf{F}_{j}$ act at positions $\mathbf{r}_{j},$ then the net torque (units: $N-m$ or $\left.l b-f t\right)$ is the sum $$ \boldsymbol{\tau}=\sum \mathbf{r}_{j} \times \mathbf{F}_{j} $$ Calculate the net torque about $O$ at $P$, assuming that a $30-\mathrm{kg}$ mass is attached at $P$ [Figure $21(\mathrm{~B})]$. The force $\mathbf{F}_{g}$ due to gravity on a mass $m$ has magnitude $9.8 m \mathrm{~m} / \mathrm{s}^{2}$ in the downward direction.
Vector Geometry
The Cross Product
The application point and direction of a force can affect the rotational motion of the object to which the force is applied. The tendency of a particular force to cause an angular acceleration of an object is quantified as the torque produced by the force. The torque τ is defined to be the vector cross product r × F, where r is the vector from the axis of rotation to the point where the force is applied. The magnitude of the torque is simply |τ| = |r||F| sinθ, where θ is the angle between r and F. Compute & compare the magnitude of the net torque about the pivot in part A to that in each case below: |τ1 - τ2|, |τ1 - τ0| = 135°, |τ = 2P4b1|, |τ1 - τl|, |τpl - 2FN|
Sri K.
In Exercises $67-70,$ the torque about the origin $O$ due to a force $\mathbf{F}$ acting on an object with position vector $\mathbf{r}$ is the vector quantity $\boldsymbol{\tau}=\mathbf{r} \times \mathbf{F}$. If several forces $\mathbf{F}_{j}$ act at positions $\mathbf{r}_{j},$ then the net torque (units: $N-m$ or $\left.l b-f t\right)$ is the sum $$ \boldsymbol{\tau}=\sum \mathbf{r}_{j} \times \mathbf{F}_{j} $$ Calculate the torque $\tau$ about $O$ acting at the point $P$ on the mechanical arm in Figure $21(\mathrm{~A})$, assuming that a 25 -newton force acts as indicated.
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