Skip to main content
added 116 characters in body
Source Link
Gert
  • 35.5k
  • 8
  • 62
  • 107

In the second case, decompose the force $\vec{F}$ into horizontal ($x$) component and a vertical ($y$) one:

$$F_x=F\cos\theta$$ And:

$$F_y=F\sin\theta$$

The $x$ component is what might overcome friction and thus cause motion acc. Newton's second.

The vertical one reduces the Normal force, caused by the weight $mg$:

$$F_N=mg-F_x=mg-F\cos\theta$$

This then gives rise to the friction force $F_f$, opposing $F_x$:

$$F_f=\mu F_N$$

If:

$$F\cos\theta > \mu(mg-F\cos\theta)$$

there will be acceleration to the right.

But there is no vertical friction at all.

Should it be the case that:

$$F\sin\theta > mg$$

there will be acceleration in the $y$ direction (upward).

In the second case, decompose the force $\vec{F}$ into horizontal ($x$) component and a vertical ($y$) one:

$$F_x=F\cos\theta$$ And:

$$F_y=F\sin\theta$$

The $x$ component is what might overcome friction and thus cause motion acc. Newton's second.

The vertical one reduces the Normal force, caused by the weight $mg$:

$$F_N=mg-F_x=mg-F\cos\theta$$

This then gives rise to the friction force $F_f$, opposing $F_x$:

$$F_f=\mu F_N$$

If:

$$F\cos\theta > \mu(mg-F\cos\theta)$$

there will be acceleration to the right.

But there is no vertical friction at all.

In the second case, decompose the force $\vec{F}$ into horizontal ($x$) component and a vertical ($y$) one:

$$F_x=F\cos\theta$$ And:

$$F_y=F\sin\theta$$

The $x$ component is what might overcome friction and thus cause motion acc. Newton's second.

The vertical one reduces the Normal force, caused by the weight $mg$:

$$F_N=mg-F_x=mg-F\cos\theta$$

This then gives rise to the friction force $F_f$, opposing $F_x$:

$$F_f=\mu F_N$$

If:

$$F\cos\theta > \mu(mg-F\cos\theta)$$

there will be acceleration to the right.

But there is no vertical friction at all.

Should it be the case that:

$$F\sin\theta > mg$$

there will be acceleration in the $y$ direction (upward).

Source Link
Gert
  • 35.5k
  • 8
  • 62
  • 107

In the second case, decompose the force $\vec{F}$ into horizontal ($x$) component and a vertical ($y$) one:

$$F_x=F\cos\theta$$ And:

$$F_y=F\sin\theta$$

The $x$ component is what might overcome friction and thus cause motion acc. Newton's second.

The vertical one reduces the Normal force, caused by the weight $mg$:

$$F_N=mg-F_x=mg-F\cos\theta$$

This then gives rise to the friction force $F_f$, opposing $F_x$:

$$F_f=\mu F_N$$

If:

$$F\cos\theta > \mu(mg-F\cos\theta)$$

there will be acceleration to the right.

But there is no vertical friction at all.