i dont understand this question, what does the motion has any effect on the forc
ID: 1683134 • Letter: I
Question
i dont understand this question, what does the motion has any effect on the force the table is exerting on the incline? kindly help in answer the whole question please. tyty In the figure below, the incline has mass M and is fastened to the stationary horizontal tabletop. The block of mass m is placed near the bottom of the incline and is released with a quick push that sets it sliding upward. It stops near the top of the incline, as shown in the figure, and then slides down again. always without friction. Find the force that the tabletop exerts on the incline throughout this motion i dont understand this question, what does the motion has any effect on the force the table is exerting on the incline? kindly help in answer the whole question please. tytyExplanation / Answer
In the figure below, the incline has mass M and is fastened to the stationary horizontal tabletop. The block of mass m is placed near the bottom of the incline and is released with a quick push that sets it sliding upward. It stops near the top of the incline, as shown in the figure, and then slides down again. always without friction. Find the force that the tabletop exerts on the incline throughout this motion. The motion of the block itself wouldn't effect the force exerted from the table, just the weight of the additional mass. If you think about other incline problems, the ycomponent of the block's weight is what effects the normal force (what we're concerned with in this problem), whereas the xcomponent accounts for its falling movement down the incline. So the M of the ramp will exert a force due to its weight on the table top. It's parallel to the table, so we don't need to break it into components: FwM=Mg. Because the M ramp is not accelerating through the table, we know the table is exerting an equal and opposite force on the ramp. This is the normal force: Fn=Mg But we also have block m, which will exert a force due to its weight: Fwm=mg. This block is set at some , so we do need to break it into components and isolate the Fwy. With variables, Fmwy=mgcos, which will equal the Fn exerted by the ramp for the same reason listed above. The tabletop must account for both the ramp and the additional block, so: Fntable=Mg + mgcos.Related Questions
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