Question

A crane’s trolley at point P in Fig. 4–63 moves fora few seconds to the right with constant acceleration, and the 870 kg load hangs on a light cable at a$5.0^o$angle to the vertical as shown. What is the acceleration of the trolley and load?

Short answer

The acceleration of the trolley and load is $0.86\mathrm{m}/{\mathrm{s}}^2$.

Step-by-step solution

Step 1. Meaning of a free body diagram (FBD)

A free body diagram is a symbolic representation of an object and all the forces acting on it. It helps solve problems where the equilibrium of forces is involved.

The forces are drawn from the centre of the object toward the direction they act.

Step 2. Statement of Newton’s second law of motion

Newton’s second law of motion states that the acceleration of a body is directly proportional to the net force acting on the body and inversely proportional to the mass of the body.Mathematically,

$a\propto \frac{F_{\mathrm{net}}}{m}$

Step 3. Identification of given data

The mass of the load is $m=870\mathrm{kg}$.

The angle made with the vertical is $\theta =5.0^{\mathrm{o}}$.

Step 4. Drawing an FBD for the load

A free body diagram for the load is drawn below. The angle $\theta$made with the vertical is exaggerated in the sketch for simplification.

The forces acting on the load are:

• Tension T along the cable
• Weight W of the load acting downward

Step 5. Resolution of the forces

The tension Tcan be resolved into its components. The vertical component is $T\cos \theta$, and the horizontal component is $T\sin \theta$.

The vertical component of the tension force is equal to the weight of the load acting downward.

role="math" localid="1645607406093" $\begin{array}{l}T\cos \theta =W\\ T\cos \theta =mg\dots \left(i\right)\end{array}$

The horizontal component of the tension force is responsible for the acceleration of the load and is equal to the pseudo force.

$T\sin \theta =ma\dots \left(ii\right)$

Step 6. Determination of the acceleration of the trolley and the load

From equation (i), the net force is given by:

$T=\frac{mg}{\cos \theta }\dots \left(iii\right)$

From equation (ii), the acceleration can be written as:

$a=\frac{T\sin \theta }{m}\dots \left(iv\right)$

Substitute equation (iii) in (iv).

$\begin{array}{c}a=\left(\frac{mg}{\cos \theta }\right)\frac{\sin \theta }{m}\\ =g\tan \theta \\ =\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\tan 5^{\mathrm{o}}\\ =0.86\mathrm{m}/{\mathrm{s}}^2\end{array}$

Thus, the acceleration is $0.86\mathrm{m}/{\mathrm{s}}^2$.