A block a force 1705 N is shown on an incline of 29". A force 45 N is applied to it horizontally. 1705N Us=0.35 Uk =0.25

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answerhappygod
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A block a force 1705 N is shown on an incline of 29". A force 45 N is applied to it horizontally. 1705N Us=0.35 Uk =0.25

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A Block A Force 1705 N Is Shown On An Incline Of 29 A Force 45 N Is Applied To It Horizontally 1705n Us 0 35 Uk 0 25 1
A Block A Force 1705 N Is Shown On An Incline Of 29 A Force 45 N Is Applied To It Horizontally 1705n Us 0 35 Uk 0 25 1 (86.92 KiB) Viewed 32 times
A block a force 1705 N is shown on an incline of 29". A force 45 N is applied to it horizontally. 1705N Us=0.35 Uk =0.25 45N 29° a) Determine if the block is in equilibrium. The block is not in equilibrium v b) Find the magnitude and direction of the frictional force on the block. The magnitude of the force is The normal force is N and it is normal to the ramp c) If the block is in motion, determine whether it's moving up or down the ramp. The maximum friction force is Fmax = moving down the ramp N and it is up the ramp
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