Asked by 1234
A 5.5-kg box is under an upward force of 45 N which makes a 60o angle with its normal. Find the magnitude of the horizontal acceleration. What is the normal force?
Answers
Answered by
Elena
Forces acting on the box:
m•g ↓ , N↑ , ∕ F
(F can be resolved into two components –
horizontal → F(x)=F•sinα ,
and vertical ↑F(y) =F•cosα ),
acceleration “a” →
y-axis: m•g =N + F(y),
N=m•g- F(y) = 5.5•9.8 - 45•sin 60º = ....
x-axis: ma =f(x) =F• cos 60º =....
m•g ↓ , N↑ , ∕ F
(F can be resolved into two components –
horizontal → F(x)=F•sinα ,
and vertical ↑F(y) =F•cosα ),
acceleration “a” →
y-axis: m•g =N + F(y),
N=m•g- F(y) = 5.5•9.8 - 45•sin 60º = ....
x-axis: ma =f(x) =F• cos 60º =....