如图所示,物体的质量m=4.4kg,用与竖直方向成的斜向右上方的推力F把该物体压在竖直墙壁上,使它静止。物体与墙壁间的动摩擦因数μ=0.5,取重力加速度g=10m/s2,最大静摩擦与滑动摩擦相等,(sin37°=0.6,cos37°=0.8)则推力F的大小可能是( ) A.22N B.44N C.66N D.98N 答案:【答案】BC【解析】当物体刚要上滑时,静摩擦力沿墙壁向下达到最大值,此时力F为最大,设为F1,根据正交分解得,联立解得当物体刚要下滑时,静摩擦力沿墙壁向上达到最大值,此时力F最小,设为F2,根据正交分解得,又f2=μN2,联立解得故推力F的范围是故选BC。
I still doubt ________ with these useless books.
[ ]
A.
how can he deal
B.
what he can do
C.
what to deal
D.
how to do