第二道!紧急A block of mass m = 2.0 kg is held in equilibrium on an incline of angle θ = 60° by the horizontal force FDetermine the normal force exerted by the incline on the block

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第二道!紧急A block of mass m = 2.0 kg is held in equilibrium on an incline of angle θ = 60° by the horizontal force FDetermine the normal force exerted by the incline on the block

第二道!紧急A block of mass m = 2.0 kg is held in equilibrium on an incline of angle θ = 60° by the horizontal force FDetermine the normal force exerted by the incline on the block
第二道!紧急
A block of mass m = 2.0 kg is held in equilibrium on an incline of angle θ = 60° by the horizontal force F
Determine the normal force exerted by the incline on the block

第二道!紧急A block of mass m = 2.0 kg is held in equilibrium on an incline of angle θ = 60° by the horizontal force FDetermine the normal force exerted by the incline on the block
一个m=2.0kg的物块放在一个60度倾角的斜面上,受到一个水平力F的作用而处于平衡状态,求斜面对物块施加的力.
物块受到3个力的作用,重力G,水平力F和斜面的力F1,三个力的合力为零,则
mg=F1*cos60° （竖直方向）
F=F1*sin60°
解得 F1=mg/cos60°=39.2N

中文回答
20N
物体沿斜面方向上受力平衡 把重力和水平力F在斜面分别分解

20kg的物体放在60度斜面上受水平力F作用
决定物体运动的最小力

质量为20kg的砖块放在60度斜面上,加一水平力F,达到平衡状态,让你确定斜面对砖块的作用力
很详细吧,求你了,我没分了!!!!!

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