Question

Two forces $$\vec{F}_{1}$$ and $$\vec{F}_{2}$$ act on a $$5.00kg$$ object. Taking $$F_{1}= 20.0 N$$ and $$F_{2}= 15.0 N$$, find the accelerations of the object for the configurations of forces shown in parts (a) and (b) of above figure.

Answer 1

We use the particle under a net force model and add the forces as vectors. Then Newton’s second law tells us the acceleration.
(a) $$\sum\vec{F}=\vec{F}_{1}+\vec{F}_{2}=(20.0\hat{i}+15.0\hat{j})N$$
Newton's second law gives, with $$m=5.00kg$$
$$\vec{a}=\frac{\sum\vec{F}}{m}=(4.00\hat{i}+3.00\hat{j})m/s^{2}$$
or $$a=5.00m/s^{2}$$ at $$\theta=36.9^{0}$$
(b) In this configuration,
$$F_{2x} = 15.0\cos60.0^{0} = 7.50 N$$
$$F_{2y} = 15.0\sin60.0^{0} = 13.0 N$$
$$\vec{F}_{2} = (7.50\hat{i} + 13.0\hat{j}) N$$
Then,
$$\sum\vec{F}=\vec{F}_{1}+\vec{F}_{2}=[20.0\hat{i}+(7.50\hat{i}+13.0\hat{j})]N$$
$$=(27.5\hat{i}+13.0\hat{j})N$$
and $$\vec{a}=\frac{\sum\vec{F}}{m}=(5.50\hat{i}+2.60\hat{j})m/s^{2}=6.08m/s^{2}=6.08m/s^{2}$$ at $$25.3^{0}$$