Một thanh chắn đường AB dài 8 m;có khối lượng 30 kg,có trọng tâm G cách đầu A là 1,5 m.Thanh có thể quay quanh một trục O nằm ngang cách đầu A là 2 m.Để giữ thanh cân bằng nằm ngang thù phải tác dụng lên đầu B một lực bằng bao nhiêu.
Lấy g=10 m/s2.
Lấy g=10 m/s2.
Quảng cáo
1 câu trả lời 232
2 năm trước
Để thanh cân bằng nằm ngang thì moment lực của lực B tác dụng lên đầu B phải bằng moment lực của trọng lực G của thanh:
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Với FB là lực tác dụng lên đầu B, AO là khoảng cách từ trục quay đến đầu A, G là trọng lượng của thanh, OG là khoảng cách từ trục quay đến trọng tâm G.
Thế giá trị cho các thông số ta có:
AO=2m, OG=GA−OG=1.5m, G=mg=30kg⋅10m/s2=300N
Vậy:
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Vậy để giữ thanh cân bằng, cần tác dụng lên đầu B một lực bằng 225 N.
Với FB là lực tác dụng lên đầu B, AO là khoảng cách từ trục quay đến đầu A, G là trọng lượng của thanh, OG là khoảng cách từ trục quay đến trọng tâm G.
Thế giá trị cho các thông số ta có:
AO=2m, OG=GA−OG=1.5m, G=mg=30kg⋅10m/s2=300N
Vậy:
Vậy để giữ thanh cân bằng, cần tác dụng lên đầu B một lực bằng 225 N.
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